6.13 Theorems on Triangles and circles:

6.13.1 Basic proportionality theorem

You must have heard about pyramids in Egypt. How did ancient people measure the height of pyramids without actually climbing them?

They used the concept of ‘similarity’ of polygons to measure the heights and distances which are otherwise difficult to find.

Similarity concept is extensively used in vehicle manufacturing, space technology and many other engineering fields

We have figures of two pentagons on the right side What do you observe? Don’t you notice that they are identical except for their sizes? Actually the figure on extreme left side is 50% smaller than the figure on its right side. When you measure the length of the sides of these figures, You will notice that they are in the same ratio. AB/ PQ = BC/ QR = CD/ RS = ……= 1/ 2 You will also notice that the angles at each of the corresponding vertices are same. EAB = TPQ, ABC = PQR, BCD = QRS , . . .

Two polygons having same number of sides are similar (denoted by the symbol ‘|||’) if and only if:

1. The corresponding angles of two polygons are equal

AND

2. The corresponding sides of two polygons are proportional (i.e. ratios of their sides are equal)

Notice that regular polygons of same number of sides are always similar.

If two rectangles are to be similar then the ratio of their lengths = ratio of their breadths as in the figure given below

Let us observe the following figures; Don’t they appear of same shape? ABC = DEF, BAC =EDF, ACB = DFE The ABC and DEF are similar triangles and we represent this by ABC ||| DEF .

	Congruent Figures	Similar Figures
1	Figures have same shape and same size.	Figures have same shape and sizes are in same proportion
2	Corresponding sides are equal	Ratios of corresponding sides are same
3	Congruent triangles are similar.	Similar triangles need not be congruent.

In 6.3.3 we have observed that it is not possible to construct unique triangle given three angles.

If two triangles have two pairs of corresponding angles equal, then triangles are similar (This is called ‘AAA (Angle, Angle, Angle) Postulate

on similarity’.

Note: The above postulate is also called ‘AA (Angle, Angle) Postulate on similarity’.(The reason is that, if two angles are equal in a triangle then

third angle has to be equal because sum of all angles in a triangle is 180⁰)

6.13.1. Basic Proportionality Theorem (Thale’s theorem) -- (BPT) A straight line drawn parallel to a side of a triangle, divides the other 2 sides

proportionately.

Data :In ABC, X is a point on AB and the line XY is ||BC,

To Prove: AX/BX = AY/CY.

Step	Statement	Reason
	Consider ABC and AXY
1	AXY = ABC	Corresponding angles(XY||BC)
2	AYX = ACB	Corresponding angles(XY||BC)
3	XAY = BAC	Common angle
4	ABC ||| AXY	AAA Postulate on similarity
5	AB/AX = AC/AY	Corresponding sides are proportional
6	(AX+BX)/AX = (YC+AY)/YC	Spilt AB and AC
7	1+BX/AX = 1+AY/YC	Simplification
8	BX/AX = AY/YC	Subtract 1 from both sides of Step 5
9	AX/BX =AY/CY	Reciprocal of 6

6.13.1 Problem 1: In the above figure if XY || BC, prove that AB/BX=AC/YC

Step	Statement	Reason
1	AX/XB=AY/YC	BPT
2	1+(AX/XB)=1+(AY/YC)	Add 1 to both sides
3	(XB+AX)/XB=(YC+AY)/YC	X and Y are points on AB and AC
4	AB/XB= AC/YC

Conversely, If a straight line divides the two sides of a triangle proportionately then that line has to be parallel to the third side.

(Note: Proof is not provided for this (The proof is by logical reasoning).

Note: Basic proportionality theorem’s concept can be used to divide a given line in to different ratios (Refer 6.1 Problem 11)

6.13.1. Corollary to BPT:

If a line is drawn parallel to a side of a triangle, then the sides of the new triangle formed are proportional to the sides of the given triangle

(New triangle is ‘similar’ to the given triangle)

Data: XY is parallel to BC

To prove: AX/AB=AY/AC=XY/BC

Construction: Draw a line parallel to AC from X and let that line meet BC at Z

Proof:

Step	Statement	Reason
1	BZ/BC=BX/BA	BPT since XZ || AC (Construction)
2	XY=ZC, YC =XZ	By construction, XZCY is a parallelogram and hence opposite sides are equal
3	AX/AB = AY/AC	BPT since XY || BC (Given)
4	LHS =AX/AB = (AB-BX)/AB	X is a point on AB
5	= 1-(BX/AB)
6	= 1-(BZ/BC)	From (1)
7	= (BC-BZ)/BC	Z is a point on BC
8	= ZC /BC
9	=XY/BC= AY/AC	From(2)

6.13.1 Problem 2: Show that in a trapezium, the line joining the midpoints of non parallel sides is parallel to the parallel sides

Data: BCED is a trapezium; X and Y are midpoints of DB and CE respectively

To prove: XY || DE

Construction: Extend DB and EC to meet at A (since DB and EC are non parallel lines, they have to meet at a point when extended)

Step	Statement	Reason
1	AB/BD = AC/CE	BPT (DE || BC)
2	(AX-BX)/BD = (AY-CY)/CE	X and Y are points on DA and EA
3	AX/BD – BX/BD = AY/CE –CY/CE	X and Y are mid points so BD=2BX=2DX,CE=2YE=2CY
4	AX/2DX -1/2 =AY/2YE -1/2	From 3
5	AX/DX =AY/YE	Add 1/2 to both sides in 4 and then multiply both sides by 2
6	XY || DE	Converse of BPT

6.13.1. Theorem 1: If two triangles are equiangular then their corresponding sides are proportional

Data: In ABC and DEF

BAC = EDF

ABC =DEF

ACB =DFE

To Prove: AB/DE = BC/EF= AC/DF

Construction: Mark X on AB and Y on AC such that AX=DE and AY=DF. Join XY

Proof:

Step	Statement	Reason
1	BAC = EDF	Given
2	AX=DE,AY=DF	Construction
3	AXY and DEF are congruent	SAS Postulate
4	AXY=DEF	Congruent triangle property
5	= ABC	Given
6	XY||BC	Corresponding angles AXY and ABC are equal
7	AB/AX=AC/AY=BC/XY	Corollary to BPT
8	AB/DE=AC/DF=BC/EF	From(2)

6.13.1. Converse of Theorem 1: If the corresponding sides of two triangles are proportional

then the triangles are equiangular

Data: In ABC and DEF

AB/DE=AC/DF=BC/EF

To Prove:

BAC = EDF

ABC =DEF

ACB =DFE

Construction: Mark X on AB and Y on AC such that AX=DE and AY=DF. Join XY

Step	Statement	Reason
1	AB/DE=AC/DF=BC/EF	Given
2	AX =DE,AY=DF	Construction
3	AB/AX=AC/AY	Step 1 and 2(Substitution)
4	XY||BC	Converse of BPT
5	AXY=ABC, AYX=BCA	Corresponding angles are equal
6	AXY=DEF, AYX=DFE	By construction AXY, DEF are congruent
7	ABC=DEF, BCA=DFE	Step 5, 6(Substitution)
8	BAC =EDF	When two angles are equal third angle has to be equal

6.13.1 Problem 3: ABC is a triangle in which AB=2cm, BC=3cm and CA = 4cm. DEF is another triangle whose largest side is 6cm. If DEF is similar to ABC,

find the relationship between perimeters of these two triangles.

Solution:

The perimeter of ABC= 2+3+4 =9cm Let the largest side in DEF be DF=6cm Since triangles are similar we have DE/AB = EF/BC =DF/AC Since DF=6 and CA = 4, DF/AC =6/4=1.5 DE = 1.5*AB = 1.5*2 = 3cm and EF =1.5*BC=1.5*3=4.5cm The perimeter of DEF = 6+3+4.5 =13.5cm which is one and half times the perimeter of ABC.

Thus, the perimeters of similar triangles are in same proportion as their corresponding sides

6.13.1 Problem 4: You must have heard about Meharuli Pillar near Qutubminar. It is an iron pillar installed by Chandragupta II who ruled northern

India around 400AD. It is amazing to note that it has not rusted till today. This pillar is quite tall. Let us use properties of similarity to find its height.

Assume that on a visit to the place, you are standing 9feet 2inches away from this pillar and you notice that the shadow cast by the pillar is

2 feet 8 inches. If your height is 5’4”. Find the height of the pillar.

Solution:

For simplicity let us convert all measures to inches. Let the pillar be represented by AB. You are at D, 110 inches (9’2”) away from the pillar. Your height is represented by DP = 64” (5’4”). Shadow cast by pillar (DC) is 32 inches. BAC =PDC = 90^0 Since AB || DP, ABP = DPC ACB is a common angle to both BAC and PDC. Therefore these two triangles are equiangular and hence their corresponding sides are proportional(AAA Postulate) Hence AB/PD = AC/DC. Note that AC =AD+DC = 110+32= 142 AB = AC*PD/DC = 142*64/32=284’’ The height of iron pillar = 284 inches = 23 feet 8inches. You will notice from history books that this answer is correct.

6.13.1 Problem 5: ABC is a right angled triangle at B and D is any point on AB. DE is perpendicular to AC. If AD=4m, AB=16m, AC=24m. Find AE

Solution:

ABC =DEA = 90^0. Since A is the common vertex of triangles ABC and DAE, BAC = DAE. Since two angles of these triangles are same, third angle also has to be same (BCA=ADE). Therefore ABC and DEA are equiangular (similar) triangles. Therefore by theorem, corresponding sides are proportional. The corresponding sides are: AC, AD (hypotenuse) BC, DE (opposite side of A) AB, AE (opposite to BCA and ADE) AE/AB = DE/BC=AD/AC AE = AB*AD/AC = 16*4/24 = 16/6 =2.67m

6.13.1 Problem 6: In the trapezium ABCD, AD||BC and the diagonals intersect at O. Prove that OB/OC = OD/OA

Solution:

Since AD||BC, BCO = OAD, CBO=ODA (alternate angles) Since the lines AC and BD intersect at O, COB =AOD (vertically opposite angles) The triangles BCO and DOA are equiangular. Hence by theorem the corresponding sides are proportional The corresponding sides are OB, OD (opposite to angles BCO and OAD) OC, OA (opposite to angles CBO and ODA) By theorem OB/OD = OC/OA i.e. OB*OA=OD*OC OB/OC = OD/OA

6.13.1 Theorem 2: Areas of similar triangles are proportional to the squares of the corresponding sides

Data: Let ABC and DEF be similar triangles, in which (BC,EF), (AB,DE) and (AC,DF) are pairs of corresponding sides.

To Prove: Area of ABC/Area of DEF = BC²/EF²= AB²/DE² =AC²/DF²

Construction: Draw AL perpendicular to BC and DM perpendicular to EF

Step	Statement	Reason
1	Area of ABC= BC*AL/2	Area of triangle = base*height/2
2	Area of DEF = EF*DM/2	Area of triangle = base*height/2
3	(Area of ABC)/ (Area of DEF)=(BC/EF)*(AL/DM)
4	ABL =DEM	Triangles ABC and DEF are similar (Given)
5	ALB =DME= 90^0	Construction
6	BAL =EDM	Since 2 angles are equal, third angle is also equal
7	AL/DM = AB/DE =BL/EM	Corresponding sides are proportional in similar triangles
8	But AB/DE =BC/EF	Triangles ABC and DEF are similar (Given)
9	AL/DM=BC/EF	From (7) and (8)
10	(Area of ABC)/ (Area of DEF)=(BC/EF)*(BC/EF)= BC^2/EF^2	Substitute (9) in (3)
11	Similarly we can prove other 2 equalities

6.13.1 Corollary : If the areas of two similar triangles are equal then they are congruent

Proof :

Let ABC and DEF be two similar triangles with BC and EF as base By above theorem (Area of ABC)/ (Area of DEF) = BC^2/EF^2 Since the areas are same it follows that BC=EF, Similarly we can prove that other corresponding sides are also equal Hence by SSS Postulate they are congruent.

6.13.1 Problem 7: Prove that the area of similar triangles have the same ratio of the squares of corresponding altitudes

Solution:

In the above theorem, in step 7, we had proved that triangles ALB and DME are equiangular. Because of this, the corresponding sides are proportional (Theorem 1) AB/DE = AL/DM (AB/DE)^2= (AL/DM)^2 Since the triangles ABC and DEF are similar, by the above theorem Area ofABC /Area of DEF = BC^2/EF^2 = AB^2/DE^2 = AL^2/DM^2 = squares of the ratio of the altitudes

6.13.1 Problem 8: In ABC, BE is Perpendicular to AC and CF is perpendicular to AB. BE and CF intersect at O.

Prove that Area ofBOF / Area of COE= BF²/CE²

Step	Statement	Reason
1	BFO =CEO = 90^0	Given
2	BOF =COE	Opposite angle
3	FBO =OCE	Since 2 angles are equal, third angle has to be equal
4	BFO ||| CEO	Triangles are equiangular
5	Area of BFO /Area of CEO = BF^2/CE^2	By theorem

6.13.1 Problem 9: In the adjoining figure ABC and DBC are two triangles on the same base. Prove that Area of ABC /Area of DEF = AO/DO

Construction: Draw Perpendicular to base BC from A and D to cut the line BC at E and F.

Step	Statement	Reason
1	AOB =DOF	Vertically opposite angle
2	AEO =DFO= 90^0	Construction
3	AEO |||DFO	AAA postulate on similarity
4	AE/DF = AO/DO	Corresponding sides are proportional
5	Area of ABC /Area of DEF = {(1/2)BC*AE}/{(1/2)BC*DF}=AE/DF	Formula for area
6	Area of ABC /Area of DEF=AO/DO	Step 4

6.13.1 Problem 10: Prove that internal bisector of an angle of a triangle divides the opposite side in the ratio of the sides containing the angle.

(This is also called vertical angle theorem)

Construction: Draw a line from C parallel to AD to meet the extended BA at E(DA||CE)

To Prove: AB/AC = BD/DC

Step	Statement	Reason
1	DA||CE	Construction
2	BD/DC = BA/AE	BPT on BCE
3	BAD =AEC	Corresponding angles DA||CE
4	DAC =ACE	Alternate angles DA||CE
5	BAD =DAC	AD is bisector of BAC
6	AEC =ACE	Step 4,5
7	AE=AC	Step 6 (CAE is isosceles)
8	BD/DC = BA/AC	Step 2

Note: Converse Of the above problem is that, if a line segment drawn from the vertex of an angle of a triangle to its opposite side divides the line

in the ratio of the sides containing the angle, then the line segment bisects the angle.

6.13.1 Problem 11: In the Adjoining figure diagonal BD of a quadrilateral ABCD bisects B and D Prove that AB*CD =AD*BC

Hint : 1. ADB = CDB, ABD = CBD( Given that BD bisects B and D) 2. BY AAA postulate on similarity ADB |||CDB 3. Hence corresponding sides are proportional: AB/BC =AD/CD AB*CD =AD*BC

6.13.1 Problem 12: What is the height of the intersecting point joined by strings tied to the base and top of poles of height 15 and 10 ‘Mola’s( A unit of measurement)? ( Lilavati Shloka 162) .

Step	Statement	Reason
1	OP/CD = BP/BD	OP || CD , BPT
2	OP/AB=PD/BD	OP || AB , BPT
3	OP/CD +OP/AB = (BP/BD)+ (PD/BD)	Add (1),(2)
4	OP(AB+CD)/(AB*CD) = (PD+BP)/BD =1	(PD+BP) =BD
5	OP = (AB*CD)/ (AB+CD) = 15*10/25 = 6
Note : we can find BP or PD given PD or BP

6.13.1 Problem 12: O Lilavati, Tell me the distance between the toy and the ‘Deepa’ ( Candle stand) and the height of ‘Deepa’ if:

the height of toy is 12” and the height of its shadow is 8” in its original place

height of the shadow becomes 12” if toy is moved forward by 2 ‘Hasta’ ( 1 Hasta = 24”)

( Lilavati Shloka 244). Bhaskara also gives formulae to solve these types of problems.

Solution : In the figure

AB is the Deepa, CL and EM are the initial and final positions of the toy whose height is 12”.

CD is the shadow of the toy at C and the length of the shadow is 8”.

New position of the toy is at E (CE = 48”( 2*24))

EF is the shadow of the toy at E its length is 12”.

Step	Statement	Reason
1	AB/LC = BD/CD	LC || AB , BPT
2	AB/ME =BF/EF	ME || AB , BPT
3	BD/CD = BF/EF	LC=ME=12, From (1),(2)
4	(BC+CD)/CD = (BC+CD+DE+EF)/12	BD = (BC+CD), BF =(BC+CD+DE+EF) , EF=12
5	(BC/8)+1 = (BC/12)+5	CD=8, (CD+DE+EF) = 8+40+12 =60
6	BC(1/8-1/12) = 4
7	BC{(6-4)/48} =4 BC =96 BD = 104	BD = BC+CD = 96+8
8	AB = (BD*LC)/CD= (104*12)/8 = 156	From (1)

6.13.2 Pythagoras Theorem:

In a right angled triangle, the square of hypotenuse is equal to sum of squares of other 2 sides.

There are around 50 proofs for this theorem and we use one of the simplest proof here.

It is generally represented as

AC² = AB²+BC² when right angled at B

Data: In triangle ABC = 90⁰

To Prove: AC² = AB²+BC²

Construction: Draw BD perpendicular to hypotenuse AC

Proof:

Step	Statement	Reason
1	ABC= 90^0	Given
2	BDA= 90^0	Construction
3	BAC =BAD	Common angle
4	ABD =BCD	Since 2 angles are equal, third angle has to be equal
5	ABC ||| ADB	Triangles are equiangular
6	AB/AD = AC/AB	Ratios of sides are in proportion(6.13.1 Theorem 1) (side opposite to C and hypotenuse)
7	AB^2 = AC*AD	Simplification
8	ABC ||| CDB	Repeat the steps 1 to 5 for ABC and CDB
9	BC/CD=AC/BC	Ratios of sides are in proportion(6.13.1 Theorem 1) (side opposite to A and hypotenuse)
10	BC^2 = AC*CD	Simplification
11	AB^2 + BC^2 =AC*AD+AC*CD =AC(AD+CD)=AC*AC= AC^2	Add (7) and (10) and note D is a point on AC

Have you noticed the following?

1. 25 =5² = 3²+4² = 9+16

2. 100 = 10² = 8²+6² =64+36

The triplets like (3,4,5), (6,8,10), (8,15,17), etc are called Pythagorean triplets

Note : It was Baudhayana (600BC), an Indian mathematician who first found this relationship in a right angled triangle !

Converse of Pythagoras Theorem: If the square of one side is equal to sum of squares of other two sides then these two sides(AB,BC) contain a right angle

(If AC² = AB²+BC² then B= 90⁰)

Proof is provided by constructing another right angled triangle and then proving that both are congruent (By using SSS postulate).

6.13.2 Problem 1: A school has a compound wall of 400 feet. A painter was asked to paint the wall. The painter charges Rs 80 per sq. ft to paint the wall.

The painter has a ladder of length 10 feet. When the ladder rests against the wall its foot is 6 feet away from the compound wall. Find out how much the

school pays the painter to paint the compound wall.

Solution:

If we find the area of the compound wall, we can multiply this area by the rate, to know howmuch the painter is to be paid for his work. We know AC^2 = AB^2+BC^2 Substituting values for AC=10 and BC=6 we get 10^2 = 6^2+AB^2 By transposition we get AB^2=100-36 =64 AB = Height of the wall = 8 feet Area of the compound wall = 400*8 = 3200 sq ft. Since the painter charges Rs 80 per sqft. Total amount payable is Rs 2,56,000 (=3200*80).

6.13.2 Problem 2: You are selected from your school to participate in chess a game being conducted in another school. You cycled to reach that

school as follows:

From your house you cycled 8 km to the north and then 5 km to east and then 4 km to the north.

How many km did you peddle to reach the other school?

Solution:

Draw a figure to represent AB= 8km, BC=5km and CD=4 km. The route you cycled is ABCD. We are required to find DA. Draw a line DE parallel to CB to meet the extended line AB at E. Since DE is parallel to BC and BE is parallel to CD, BCDE is a rectangle DE= 5 and AE = AB+BE = 8+4 =12 Note that DEA is a right angled triangle AD^2 = ED^2+EA^2= 5^2+12^2= 25+144 = 169 = 13^2 AD= 13km

6.13.2 Problem 3: AD is the altitude through A in the ABC and DB:CD = 3:1. Prove that BC² = 2(AB²-AC²)

Step	Statement	Reason
1	BD^2 = AB^2-AD^2	Pythagoras theorem on ABD
2	AC^2 = AD^2+CD^2	Pythagoras theorem on ADC
3	AD^2 = AC^2-CD^2	Step 2
4	BD^2 = AB^2-(AC^2 -CD^2) = AB^2-AC^2 +CD^2	(1) and (3)
5	DB/CD = 3 BC = BD+CD =4CD	Given
6	DB=3CD BD^2=9CD^2	Step 5
7	9CD^2 = AB^2-AC^2 +CD^2	(6) and (4)
8	8CD^2 = AB^2-AC^2	Transposition
9	16CD^2 = 2(AB^2-AC^2)	Multiply 8 by 2
11	BC^2= 2(AB^2-AC^2)	Step 5

6.13.2 Problem 4: A Bamboo pole of height 32 ’Mola’(Unit of measurement : say feet) broke due to heavy wind and fell on the ground. If the pole’s top touched the ground at 16 ‘Mola’ away from its base, find the point at which it broke.(Lilavati Sholka 150)

Solution:

AB is the pole of height 32 feet.Let the pole break at D to touch the ground at C BC = 16 feet. Note AB= BD+DC = 32 feet By Pythagoras theorem DC^2 = DB^2+BC^2= DB^2+16^2= DB^2+256 Let BD = x and DC = y So we have two equations to solve x+y =32 -----à(1) And y^2 = x^2+256 -----à(2) From (1) we get y = 32-x By substituting this value of y in (2) we get (32 - x)^2= x^2+256 LHS = 32^2+ x^2- 64x (expand (a-b)^2= a^2+ b^2-2ab) = 1024 + x^2- 64x Since LHS= RHS 1024 + x^2- 64x = x^2+256 768 = 64x (Transposition) x = 12 The pole breaks at 12feet from the bottom. Verification: since x=12, y = 32-12 = 20 We observe that 400 = 20^2 = 144+256 = 12^2+16^2 This is as per Pythagoras theorem.

6.13.2 Problem 5: A monkey climbed down from a tree of height 100 ‘Mola’( A unit of measurement like feet) and went to a pond 200 ‘Mola’ away. Another monkey jumped up a little distance and then jumped down diagonally and reached the same pond. If both monkeys travelled the same distance, tell me the height to which the second monkey jumped?( Lilavati : Shloka 157)

BD represents the tree of height 100 Mola(say Feet) C is the pond 200 Mola(Feet) away. Let x be the height to which the monkey jumped from the top

of tree BD. Distance travelled.

Solution:

BD represents the tree of height 100 Mola(say Feet). C is the pond 200 Mola(Feet) away. Let x be the height to which the monkey jumped from the top of tree BD.(DA=x) Distance travelled by the first monkey = DB+BC = 300 Distance travelled by the second monkey = DA+AC = x+AC It is given that distance travelled by both monkeys are same x+AC=300 AC= 300-x AC^2 = (300-x)^2=90000-600x+ x^2 ------- (1) By Pythagoras theorem AC^2 = BA^2+BC^2= (100+x)^2+200^2= 10000+200x+x^2+40000 ---------(2) From (1) and (2) 10000+200x+x^2+40000 =90000-600x+ x^2 800x = 40000 x=50 Verification : AC^2 = (100+50)^2+ 200^2 = 22500+40000 = 625000 = 250^2 100+200 = 50+ 250

6.13.2 Problem 6: A peacock is sitting on the top of a pillar of height 9 ‘Mola’(feet). An ‘ant hill’ exists at the base of the pillar. To hide itself from the attack of peacock, a snake is approaching the ‘ant hill’ from a distance three times the height of the pillar. Seeing the snake approaching the ‘ant hill’, peacock flies down to catch the snake. If both snake and peacock travel at the same speed, find the distance from ‘ant hill’ at which peacock catches the snake. ( Lilavati : Shloka 152)

Solution:

AB is the pillar of 9 feet. Peacock is sitting at A. D is the position of snake at a distance of 27 feet from the pillar. Peacock follows the path (AC) of a triangle to catch the snake. Let C be the place where peacock catches the snake. Let BC=x. CD=27-x Since both peacock and snake travel at same speed AC =CD AC=27-x AC^2 = (27-x)^2=729-54x+ x^2------- (1) From Pythagoras theorem AC^2 = BA^2+BC^2= (9)^2+x^2= 81+x^2 ---------(2) From (1) and (2) 729-54x+ x^2 =81+x^2 729-81 = 54x x= 648/54=12 Verification :: AC^2 = (27-12)^2 = 9^2 + 12^2 ( 225 = 81+144)

6.13.2 Problem 5: A circus group has put up a tent around a central pole of height of 11 meters. At 12 meters distance, from the foot of the central pole,

the circus group has put up few poles of height 6meters on the ground in a circular fashion around the central pole. The poles are strengthened by ropes

tied between the top of central pole and top of the other poles around it. Find the length of the rope required to tie the poles at the top.

Solution:

The erection of poles can be represented as in the adjoining figure. BD is the central pole of height = 11 Meters AC is one of the supporting poles of height 6 meters, put up on the ground, around the central pole. The distance between the central pole(BD) and the supporting pole(AC) on the ground (AB) = 12 meters We are required to find DC. From C, draw a line parallel to AB to cut BD at E. ABEC is a rectangle with CE=12 and BE=AC=6 ED=5 CED is a right angled triangle Hence by Pythagoras theorem DC^2 = CE^2+ED^2= 12^2+5^2=144+25 =169 = 13^2 CD = 13 meters Therefore the length of the rope required to tie these 2 poles at the top = 13 meters.

6.13.2 Problem 6: In the adjoining figure ACB = 90⁰ AB=c, BC=a, AC= b, CDB = 90⁰and CD =p

Prove that 1/p² = 1/a² + 1/b²

Step	Statement	Reason
1	AB^2 = AC^2+BC^2	Pythagoras theorem for ACB
2	c^2 = b^2+a^2	Substitution
3	(1/2)AB*CD = (1/2)cp	Area of ACB with base = AB
4	(1/2)AC*BC = (1/2)ba	Area of ACB with base = AC
5	(1/2)cp = (1/2)ba	Both areas are same
6	c = ab/p	Step 5
7	a^2b^2/ p^2 = b^2+a^2	Substitute value of c in step 2
8	1/ p^2= (b^2+a^2)/ a^2b^2	Divide both sides by a^2b^2
9	1/ p^2= 1/a^2+1/b^2

6.13 Summary of learning

No	Points to remember
1	A straight line drawn parallel to a side of a triangle divides the other 2 sides proportionately (BPT)
2	If two triangles are equiangular then their corresponding sides are proportional
3	Areas of similar triangles are proportional to the squares of their corresponding sides
4	In a right angled triangle, the square of hypotenuse is equal to sum of squares of other 2 sides

Additional points:

6.13.1 Theorem 3: Areas of triangles having a common vertex and having bases along the same line, are proportional to the lengths of their bases.

In the adjoining figure triangles ABP,ABC and ABD have the same vertex A and their bases are on the same line BD.

We are required to prove that

Area of ABP/Area of ABC = BP/BC, Area of ABC/ Area of ABD = BC/BD

Proof:

Since all the triangles have the same vertex and their bases are on the same straight line, the heights of all the triangles are same. We know that Area of a triangle = ½*Base*Height. Area of ABP/Area of ABC = (1/2*BP*height)/(1/2*BC*height) = BP/BC Similarly, Area of ABC/ Area of ABD = BC/BD

6.13.1 Theorem 4: A perpendicular drawn from the vertex of the right angle of a right angled triangle divides the triangle into two similar triangles and

they are also similar to the given triangle.

Given: In the right angled ABC, ABC = 90⁰, BDAC

To prove: ABC ||| ADB ||| BDC

Step	Statement	Reason
	Consider ADB and ABC
1	ADB = ABC = 90^0	BDAC
2	BAD = BAC	Common
3	ABC ||| ADB	AAA Postulate
	Consider BDC and ABC
4	BDC = ABC = 90^0	BDAC
5	BCD = BCA	Common
6	ABC ||| BDC	AAA Postulate
7	ABC ||| ADB |||BDC	Steps 3 and 6

Note: In the above diagram prove that BD² = AD*DC

Since ADB ||| BDC, AD/BD =BD/DC (Corresponding sides of similar triangles are proportional)

AD*DC = BD²

6.13.1 Theorem 5 (30⁰-60⁰-90⁰ triangle theorem) : If the angles of a triangle are 30⁰, 60⁰ and 90⁰ then the side opposite to 30⁰ is half the

hypotenuse and the side opposite to 60⁰ is /2 times the hypotenuse.

Given: In ABD, ABD = 60⁰, BAD = 30⁰, ADB = 90⁰

To prove: BD = AB/2, AD = /2 * AB

Construction: Produce BD such that DC = BD. Join AC.

Step	Statement	Reason
	Consider ADB and ADC
1	ADB = ADC = 90^0	Given that ADB = 90^0 and Construction
2	BD = DC	Construction
3	AD is common
4	ADB ||| ADC	SAS Postulate
5	BAD = CAD	Corresponding angles of similar triangles are equal
6	BAC = 60^0	Step 5 and it is given that BAD = 30^0
7	AB = AC = BC	ABC is an equilateral triangle (all angles = 60^0)
8	BD = DC = AB/2	Step 2, 7
9	AB^2 = AD^2+BD^2	Pythagoras theorem
10	AB^2 = AD^2+(AB/2)^2	Step 8
11	AD^2= AB^2-(AB/2)^2= (3AB^2)/4	Simplification
12	AD = /2*AB

6.13.3 Postulates on similarity:

In addition to AAA postulate on similarity discussed in the beginning of this section, we have three more postulates on similarity.

2. ‘Two triangles are similar if two of their sides are proportional and their included angles are same’ This statement is called ‘SAS (Side, Angle, Side) Postulate on Similarity’. In the adjoining figure AB/DE = BC/EF and their included angle ABC = DEF. There fore ABC ||| DEF.
3. ‘Two triangles are similar if the sides of one triangle are proportional to the corresponding sides of another triangle’. This statement is called ‘SSS (Side, Side, Side) Postulate on Similarity’. In the adjoining figure DE/AB = EF/BC = DF/AC There fore ABC ||| DEF.

Table: Postulates for similarity of triangles:

Proportional			Equal			Similarity Postulate
Side	Side	Side	Angle	Angle	Angle
-	-	-	y	y	y	AAA
Y	Y	-	Y(Included)	-	-	SAS
Y	y	y	-		-	SSS

6.13.3 Intercept Theorem: If three or more lines are intercepted by two transversals, the intercepts made by them on the transversals are

proportional. (Note this is a variation of the theorem 6.8.7)

Given: Transversal p makes intercepts on three lines, l, m and n.

(I.e. l || m || n) q is another transversal which makes intercepts DE and EF

To prove: AB/BC=DE/EF

Construction: Draw AS and BT || the line q

Steps	Statement	Reason
	Consider ABS and BCT
1	ABS = BCT, BAS = CBT,	Corresponding angles (It is given that line l || line m)
2	ABS ||| BCT	AA Postulate On similarity
3	AB/BC=AS/BT	Corresponding sides are proportional
4	ASED is a parallelogram	AS||DE(construction), AD||BE(Given)
5	AS=DE	Sides of parallelogram
6	BTFE is a parallelogram	BT||EF(construction), BE||CF(Given)
7	BT=EF	Sides of parallelogram
8	AB/BC = AS/BT = DE/EF	Step 3,5 and 7

6.13.3 Problem 1: In the adjacent figure the medians IB and JC meet at G. Prove that BG = 2GI.

Hint: Since BI and CJ are medians, by definition of a median BJ=JA and CI=IA. Hence J and I are mid points of the sides of the triangle. By mid point theorem (6.7.7) 2IJ = BC and IJ || BC Using the fact that IJ || BC and AAA postulate on similarity prove that GIJ and GBC are similar IJ/BC = GI/BG Since 2IJ = BC, it follows that GI/BG =1/2 i.e. 2GI = BG

6.13.3 Problem 2: As shown In the figure given below, M is the mid point of BC of parallelogram ABCD. DM intersects the diagonal AC at P and

it meets AB produced to E. Prove that PE = 2PD

Given: ABCD is a parallelogram; M is mid point of BC

To prove: PE = 2PD

Step	Statement	Reason
	Consider MCD and MBE
1	CM = MB	Given
2	DMC = BME	Vertically opposite angles
3	DCM = MBE	Alternate angles AB||CD
4	MCD MBE	ASA Postulate
5	DM = ME	Corresponding sides of congruent triangles are equal
	Consider ADP and MCP
6	APD = MPC	Vertically opposite angles
7	ADP = PMC	Alternate angles AD||BC
8	DAP = PCM	Alternate angles AD||BC
9	ADP ||| MCP	AAA Postulate
10	AD/CM = PD/PM	Corresponding sides of similar triangles are proportional (6.13.1 Theorem 1)
11	AD/CM = 2	AD=BC=2CM (Given)
12	PD/PM = 2	Steps 10 and 11
13	PE = PM+ME = PM+DM = PM+DP+PM = 2PM+PD	ME=DM as per step 5
14	PE = PD+PD = 2PD	2PM=PD(step 12)

6.13.3 Problem 3: In the given figure ABC, DE||BC and AD:DB = 3:5. Calculate the ratio of :

(i)AD/AB, AE/AC, DE/BC

(ii) Area of DEO/Area of EOC

(iii) Area of DBE/Area of DBO

(iv) Area of ADE/trapezium DECB

Note:

1. DOE and EOC have bases on the same line (DC) and same vertex E. Hence their areas are proportional to their bases (=DO/OC).

(Refer 6.13.1 Theorem 3)

2. DBE and DBO have bases on the same line (BE) and same vertex D. Hence their areas are proportional to their bases (=BE/BO).

3. ABC and ADE are similar triangles and hence their areas are proportional to the squares of their corresponding sides (=BC²/DE²).

(Refer 6.13.1 Theorem 2)

Steps	Statement	Reason
1	AB/AD = AC/AE = BC/DE	DE||BC and BPT Theorem(6.13.1)
2	AB/AD = (AD+BD)/AD = 1+BD/AD = 1+ 5/3 = 8/3	Given AD:DB = 3:5
3	EDO = OCB & DEO = OBC	DE||BC and Alternate angles
4	EDO ||| OBC	AAA postulate on similarity
5	DE/BC = DO/OC	corresponding sides are proportional
6	DE/BC = OE/OB	Follow above steps to prove this.
7	DE/BC = AD/AB = 3/8	Step 1, Step 2
8	Area of DEO/Area of EOC= DO/OC	Note 1
9	=3/8	Step 5,7
10	Area of DBE/Area of DBO = BE/BO	Note 2
11	=(BO+OE)/BO = 1+OE/BO = 1+(3/8) = 11/8	Step 6
12	Area of trapezium DECB = Area of ABC – Area of ADE
13	Area of trapezium DECB/ Area of ADE = (Area of ABC- Area of ADE) / Area of ADE = Area of ABC/ Area of ADE-1 = (BC/DE)^2-1 = (64/9)-1 = 55/9	Note 3

6.13.2 Problem 7: Find the area of an isosceles triangle whose equal sides is ‘a’ units and base is ‘b’ units

Solution:

In the adjoining figure let AC=BC=a and AB= b We know AD=DB =b/2 (perpendicular from vertex C bisects the base (AB)) By Pythagoras theorem AC^2= AD^2+DC^2 DC^2= AC^2- AD^2= a^2-(b/2)^2= (4a^2-b^2)/4 i.e. DC = SQRT(4a^2-b^2)/2 Area of ADC = (1/2)*base*height = (1/2)*(b/2)*SQRT(4a^2-b^2)/2 = 1/8*b*SQRT(4a^2-b^2) Area of ABC = 2*Area of ADC = ¼*b*SQRT (4a^2-b^2)

6.13.2 Problem 8: Find the area of an equilateral triangle whose equal sides are ‘a’ .

Solution:

Since all sides are equal in an equilateral triangle we can substitute b=a in the formula

arrived in the above problem (6.13.2 Problem 7).

Note that DCB = 30⁰, DBC= 60⁰, AD=DB=a/2

Area of an equilateral triangle = 1/4 *a* SQRT (4a²-a²) = (/4)a²

Using Pythagoras theorem we can prove that DC = (a/2)

We may observe that in triangle CDB the angles are in the ratio of 30:60:90 and the corresponding sides are in the ratio of (½):(/2):1 or 1::2

6.13.2 Problem 9: The sum of squares of 2 sides of a triangle is equal to twice the square of half the third side plus

twice the square of the median which bisects the third side. (This is also called as Apollonius theorem)

Given: ABC is a triangle, AD is the median

To prove: AB² + AC² = 2BD²+2AD²

Construction: Draw AE BC

Hint: Apply Pythagoras theorem to three triangles (AEB, AED, AEC). Then substitute and simplify the terms and also use the given data that BD=DC.

In section 6.4 we studied many concepts such as Incenter, Circumcenter, Orthocenter and Centroid. We shall prove their special properties here.

6.13.4 Concurrency theorems:

6.13.4 Theorem 1: Prove that angular bisectors of a triangle are concurrent.

Given: In ABC, the bisectors of ABC and BCA meet at I. IA is joined.

To prove: AI bisects BAC.

Construction: Draw IDBC, IEAC and IFAB

Step	Statement	Reason
1	ID=IF	BDI BFI (ASA Postulate with BI as common side)
2	ID=IE	CDI CEI (ASA Postulate with CI as common side)
3	IE=IF	Steps 1,2
4	AFI AEI	RHS Postulate (hypotenuse AI is common )
5	FAI=EAI	Corresponding angles are equal (Step 4)
6	AI bisects BAC	Step 5

Thus angular bisectors are concurrent at I (Incenter).

6.13.4 Theorem 2: Prove that perpendicular bisectors of a triangle are concurrent.

Given: In ABP, perpendicular bisectors SD of BP and SE of AP meet at S and SF is Perpendicular to AB.

To prove: SF is perpendicular bisector of AB.

Construction: Join SA, SB and SP.

Step	Statement	Reason
1	SB=SP	BDS PDS (Given BD=DP) and SAS Postulate with SD as common side
2	SP=SA	PES AES (Given PE=AE) and SAS Postulate with SE as common side
3	SB=SA	Steps 1,2
4	AFS BFS	Step3, (Given AFS = BFS = 90^0) and RHS Postulate with SF as common side
5	AF=BF	Corresponding sides of congruent triangles are equal(Step 4)
6	SF bisects AB	Step 5

Thus perpendicular bisectors are concurrent at S (Circumcenter).

6.13.4 Theorem 3: Prove that altitudes of a triangle are concurrent.

Given: In ABC, altitudes AD from vertex A and BE from vertex B meet at O. CO is produced to meet AB at F.

To prove: CF is perpendicular to AB.

Construction: Through A, B and C draw parallel lines to the bases BC, CA and AB respectively forming PQR.

Step	Statement	Reason
1	BCQA is a parallelogram	By construction BA||CQ, BC||AQ
2	BC=AQ	Step 1
3	BCAR is a parallelogram	By construction RA||BC, RB||AC
4	RA=BC	Step 3
5	AQ=RA (A is mid point of RQ)	Steps 2 and 4
6	AD is perpendicular to RQ	Given : AD BC and by construction BC||RQ
7	AD is perpendicular bisector of RQ	Steps 5 and 6
8	BE is perpendicular bisector of RP	Repeat steps 1 to 7 with respect to parallelograms RBCA, ABPC and BERP
9	Perpendicular bisectors AD and BE of PQR meet at O	Steps 7 and 8
10	CF is perpendicular bisector of PQ and it passes through O	Previous theorem (6.13.1 Theorem 2) in view of Step 9
11	CFAB	By construction PQ||BA

Thus Altitudes are concurrent at O (Orthocenter).

6.13.4 Theorem 4: Prove that medians of a triangle are concurrent and their division is in the ratio of 2:1.

Given: In ABC, the medians BE and CF meet at G. AG produced to meet BC at D.

To prove: BD=DC and AG:GD=BG:GE=CG:GF=2:1

Construction: Produce AD to K such that AG=GK. Join BK and CK.

Step	Statement	Reason
	Consider ABK
1	BF=FA	Given
2	GK=AG	By construction
3	FG||BK, FG=(1/2)BK	Mid Point theorem
	Consider ACK
4	EG||CK, EG=(1/2)CK	Mid Point theorem(Similar to steps 1,2)
5	GC||BK	Step 3
6	BG||KC	Step 4
7	BGCK is a parallelogram	Steps 5 and 6
8	The diagonals BC and GK bisect each other	Step 7 (Property of parallelogram)
9	BD=DC	Step 8
10	GD = DK = (1/2)GK	Step 8
11	= (1/2)AG	Step 2
12	2GD=AG	Steps 10, 11

Similarly we can prove that 2GE=BG and 2GF=GC.

Note: In case of an equilateral triangle, Centroid (G),Incenter (I),Circumcenter(C, S) and Orthocenter (O)are same 1. BD=DC, CE=EA, BF=FA (G is the point where the three medians meet, G=G) 2. ADBC, BEAC, CFAB (G is the point where the three altitudes meet, G=O) 3. AD, BE, CF are perpendicular bisectors (G is the point where the three perpendicular bisectors meet, G=S) 4. BAD =CAD, CBE=ABE, BCF=ACF (G is the point where the three angular bisectors meet, G=I)

6.13.5 Relationship between similarity and size-transformation.

Introduction: You must have seen models of houses, flats, buildings, townships…They are all scaled down versions of actual constructions. These

miniature versions give us a bird’s eye view of actual constructions. These models are ‘reduction’s of actuals. Similarly enlarged models are used

when the actual constructions to be built are very small.

Let us study the adjacent figure. ABC is a triangle and P is an external point. Points A1, B1and C1are on the lines PA, PB and PC respectively such that PA1 = (1/2)PA, PB1 = (1/2)PB, PC1= (1/2)PC and Points A2, B2and C2are on the lines PA, PB and PC respectively such that PA2 = 2PA, PB2 = 2PB, PC2 = 2PC By joining A1, B1, C1and A2, B2, C2we get A1B1C1 and A2B2C2 as two other triangles. We observe the following properties about the triangles ABC, A1B1C1 and A2B2C2: Their sides are proportional They are all similar triangles A1B1C1 is half the size of ABC A2B2C2 is twice the size of ABC (We can also prove the above observations) We define A2B2C2as the triangle ‘enlarged’ by a ‘scale factor of 2’ with respect to P. Here P is called as the ‘center of enlargement’ We define A1B1C1 as the triangle ‘reduced’ by a ‘scale factor of 1/2’ with respect to P. Here P is called as the ‘center of reduction’ The type of enlargement or reduction discussed above is called ‘Size Transformation’. Size transformation is not restricted to triangles and they are applicable for all types of geometrical figures. Properties of size transformations: The shape of the enlarged/reduced figure is same as that of the original figure If the scaling factor is k then each side of the resulting figure (image) is k times the corresponding side of the given (original) figure. If (i) k > 1 then the transformation is enlargement (ii) k < 1 then the transformation is reduction (iii) k = 1 then there is no transformation(same as original)

6.13.5 Problem 1: In the following figure, triangles APQ and AMN are the images of the triangle ABC under enlargement and reduction respectively

with A as center.

Given that MN=4cm, AB=9cm, BC=8cm, AC=6cm and CQ=3cm,

1. Name the images of B in each case

2. Find the scaling factor in each case

3. Find AN, AM, BP and PQ

Solution:

Step	Statement	Reason
1	ANM = ABC = APQ	MN||BC||PQ (Alternate angles)
2	The images of B are N and P	Step 1
3	ANM is reduction APQ is enlargement
4	MN/BC = 4/8 = ½ (scaling factor) for ANM	MN = 4 and BC = 8
5	AQ/AC = 9/6 = 1.5 (scaling factor) for APQ	AC = 6 and CQ = 3
6	1/2 = AN/AB = AN/9 --à AN = 4.5cm
7	1/2 = AM/AC = AM/6 --à AM = 3cm
8	1.5 = PQ/BC = PQ/8 --à PQ = 12cm
9	1.5 = AP/AB = AP/9 --à AP = 13.5cm
10	13.5 = AP = AB+BP = 9+BP --à BP = 4.5cm