The perpendicular drawn from the vertex of a triangle to its opposite side is called ‘altitude’.

In the adjoining figure C is the vertex. CM is perpendicular to the opposite side AB (extended).

CM is called Altitude.

Similarly we can draw 2 altitudes from other vertices

(vertex A to the opposite side BC and vertex B to the opposite side AC).

Note: Triangles have three altitudes.

Three or more straight lines are said to be ‘concurrent’ if they all pass through a common point.

This common point is called ‘point of concurrency’.

In the adjoining figure, lines AB, CD, EF and GH pass through a common point (intersection) O.

Hence all these four lines are concurrent and O is the point of concurrency.

Step 1: Construct a triangle of given measurements (say AB =7.5cm, AC=4cm, BC =7cm)

Step 2: With vertex C as center, draw an arc of suitable radius to cut the opposite side AB at

 2 points (if necessary extend the side) X, Y as shown in the figure.

Step 3: With X and Y as centers, draw arcs (of radius more than half of XY) to cut at point Z.

Step 4: Join C, Z to intersect the opposite side at L, CL is the altitude.

Step 5:  With Vertex A as center draw an arc of suitable radius to cut the opposite side BC at

2 points (if necessary extend the side) H, G as shown in the figure.

Step 6: With H and G as centers draw arcs (of radius more than half of HG) to cut at point I.

Step 7: Join A, I to intersect the opposite side at N, AN is an altitude.

Step 8:  With Vertex B as center draw an arc of suitable radius to cut the opposite side AC

 at 2 points (if necessary extend the side) E, D as shown in the figure.

Step 9: With E and D as centers draw arcs (of radius more than half of ED) to cut at point F.

Step 10: Join B, F to intersect the opposite side at M, BM is an altitude.

We notice that all the three altitudes  intersect at a common point O.

The point of concurrence of three perpendiculars drawn from the vertices of a triangle to their opposite sides is called ‘orthocenter’ and is usually denoted by ‘O’ 

In figure 1, ABC is an obtuse angled triangle and its orthocenter O, is outside the triangle.

In figure 2, ABC is an acute angled triangle and its orthocenter O, is inside the triangle.

In figure 3, ABC is a right angled triangle and its orthocenter O, is vertex of the right angle itself.

The line joining a vertex of the triangle to the middle point of the opposite side is called ‘median’.

In the adjoining figure of triangle ABC

A, B, C are the three vertices. L, M, N are midpoints of AB, BC and AC respectively.

The line CL joining the vertex (C) to the mid point (L) of opposite side AB is called median.

The line AM joining the vertex (A) to the mid point (M) of opposite side BC is called median.

The line BN joining the vertex (B) to the mid point (N) of opposite side AC is called median.

Step 1: Construct a triangle of given measurements (say AB = 5cm, AC = 5cm and CAB = 50⁰).

Step 2: Bisect side AB (Draw arcs of radius more than half the length of AB on both sides of AB).

Let these arcs meet at X and Y.

Step 3: Join X and Y to cut the line AB at L. Join CL. CL is the median.

Step 4: Bisect side BC (Draw arcs of radius more than half the length of BC on both sides of BC).

Let these arcs meet at P and Q.

Step 5: Join P and Q to cut the line BC at M. AM is the median

Similarly construct the median from Vertex B.

Note: All the medians pass through a common point G.

Definition:

The point of concurrence of the three medians of a triangle is called ‘centroid’ and is usually denoted by ‘G’.

Measure the distance of G from the vertex and center of opposite side. We notice the following:

In  figure I (Acute angled triangle), 2GH = AG, 2GI= BG, 2GJ=GC.

In  figure II (Right angled triangle), 2GX = DG, 2GY=EG, 2GZ=GF.

In  Figure III (Obtuse angled triangle), 2GS=PG, 2GT=QG, 2UG=GR

Step 1: Construct a triangle of given measurements (AB=7.5cm, ABC=45⁰ ,AC=4cm)

Step2: With A,B as centers draw arcs of radius more than half the length of AB on both sides of AB.

Let these arcs meet at X and Y.

Step3: Join XY to meet AB at L. (Note that XY bisects AB and is perpendicular to AB)

Step4: With B,C as centers draw arcs of radius more than half the length of BC on both sides of BC.

Let these arcs meet at P and Q

Step 5: Join PQ to meet BC at M.

(Note that PQ bisects BC and is perpendicular to BC)

Step6: With A,C as centers draw arcs of radius more than half the length of AC on both sides of AC.

Let these arcs meet at T and U.

Step 7: Join TU to meet AC at N.

(Note that TU bisects AC and is perpendicular to AC)

Note that three perpendicular bisectors of a triangle are concurrent and they meet at S.

Definition:

The point of concurrence of the perpendicular bisectors of the sides of a triangle is called ‘Circmcenter’ and is normally denoted by ‘S’ or ‘C’.

Figure 1

Figure 2 

Figure 3

In figure 1, PQR is an obtuse angled triangle and its circumcentre S is outside the triangle.

In figure 21, ABC is an acute angled triangle and its circumcentre S is inside the triangle.

In figure 32, DEF is a right angled triangle and its circumcentre S is on the hypotenuse

Step1: construct a triangle ABC with the given measurements.

Step2: With A as center, draw arcs of same radius to cut the sides AB and AC at P and Q respectively.

Step3: With P and Q as centers and with more than half of PQ as radius, draw arcs to intersect at R.

Step4: Join AR, this is the angular bisector of CAB.

Step5: With B as center, draw arcs of same radius to cut the sides BC and BA at T and S respectively

Step6: With T and S as centers and with more than half of TS as radius, draw arcs to intersect at U.

Step7: Join BU, this is the angular bisector of ABC.

Step8: With C as center, draw arcs of same radius to cut the sides CA and CB at W and V respectively.

Step9: With W and V as centers and with more than half of WV as radius, draw arcs to intersect at X.

Step10: Join CX, this is the angular bisector of ACB

Step 11: Mark the point of concurrence as I.

Note that three angularperpendicular bisectors of a triangle are concurrent and they meet at I.

Definition:

The point of concurrence of three angular bisectors of a triangle is called ‘Incenter’ and is usually denoted by ‘I’.

Step1: Construct incenter for the triangle as described above

Step 2: From I, draw perpendiculars to each of the 3 sides -AB, BC and CA and let they meet these sides at L,M and N respectively.

Notice that IL=IM=IN.

Step 3: With I as center draw a circle of radius IL.

Definition:

The circle with Incenter of the triangle as the center and which touches the sides of the triangle is called ‘incircle’.

Concurrent Lines

Point of concurrence

Name

Altitudes

O

Orthocenter

Angular bisectors

I

Incenter

Medians

G

Centroid(Center of gravity)

Perpendicular bisectors

S/C

Sircumcenter

Point of concurrence

Type of triangle

Location of point of concurrence

Centroid (G)

Any type of triangle

Inside the triangle

In center(I)

Any type of triangle

Inside the triangle

Circumcentre(C,S)

Acute angled triangle

Inside the triangle

Right angled triangle

Mid point of hypotenuse

Obtuse angled triangle

Outside the triangle

Orthocenter(O)

Acute angled triangle

Inside the triangle

Right angled triangle

Vertex of right angle

Obtuse angled triangle

Outside the triangle

As discussed above every triangle will have four concurrent points.

If that is the case. given a triangle shaped wooden object as shown in the adjacent figure, will it be possible to suspend the object in the air using a string passing through a point maintaining equilibrium status?

If possible which is that point?

No

Points to remember

1

The perpendicular drawn from the vertex of a triangle to its opposite is called altitude.

2

Point of concurrency is a point through which, three or more concurrent lines pass through

3

Orthocenter(O) is the point of concurrency of the altitudes of a triangle

4

Centroid(G) is the point of concurrency of the medians of a triangle

5

Circmcenter(S) is the point of concurrency of the perpendicular bisectors of a triangle

6

Circumcircle is a circle which passes through all the vertices of a triangle and its center is Circmcenter(S)

7

In circle is the circle  which touches internally all the sides of a triangle and its center is  in center(I) which is point of concurrence of angular bisectors.