6.5 Concurrent lines of triangles:
Definition:
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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. |
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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. |
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Activity: Construct few triangles and draw altitudes
from each of the vertex of the triangle. Did you observe that altitudes are
concurrent lines?
6.5.1 Construction of Altitudes:
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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. |
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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. |
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We notice that all the three altitudes
intersect at a common point O. |
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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’ |
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Let us observe the
position of orthocenter in three different types of triangles (obtuse, acute
and Right angled triangle).
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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. |
Definition:
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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. |
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6.5.2 Construction of
Medians:
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Step 1: Construct a triangle of
given measurements (say AB = 5cm, AC = 5cm and 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. |
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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. |
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Note: All the medians pass through a common point G. |
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Definition: The point of concurrence of the three medians of a triangle is called ‘centroid’ and is usually denoted by ‘G’. |
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Let us look at the property of centroid for different types
of triangles (Acute, Right and Obtuse angled triangles).
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Measure the distance of G from the vertex and center of opposite side. We notice the following: |
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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 |
Thus we
conclude that the Centroid divides the median in the ratio of 2:1 with respect to
the vertex and opposite side.
Definition: The bisector of any side of a triangle is called ‘perpendicular bisector’.
6.5.3 Construction of
perpendicular bisector
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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) |
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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) |
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Note that three perpendicular
bisectors of a triangle are concurrent and they meet at S. |
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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’. |
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We observe that any point on the perpendicular bisector XY
is equidistance from both A and B, and hence the point S is equidistance from
both A and B.
Similarly any point on the perpendicular bisector PQ is
equidistance from both B and C, and hence the point S is equidistance from both
B and C.
So we observe that SA=SB=SC.
If we draw a circle with S as center and SA as radius, we
see that the this circle passes through all the vertices (A, B, C) of the
triangle ABC.
Definition:
A circle which passes through all the vertices of a
triangle is called ‘circumcircle’ of the triangle.
Let us look at the position of circumcentre in three different
types of triangles (acute, right and obtuse angled triangle).
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Figure 1 |
Figure
2 |
Figure 3 |
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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 |
6.5.4 Construction of
angular bisector:
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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 |
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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 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 Step 11: Mark the point of
concurrence as I. |
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Note that three angularperpendicular
bisectors of a triangle are concurrent and they meet at I. |
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Definition: The point of concurrence of
three angular bisectors of a triangle is called ‘Incenter’ and is usually denoted by ‘I’. |
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6.5.5 Construction of
Incircle:
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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. |
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Definition: The circle with Incenter of the triangle as the center and which touches the sides of the triangle is called ‘incircle’. |
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Note: In the case of an
equilateral triangle,
Its orthocenter= it’s Incenter = it’s centroid= it’s
Circumcenter (O=I=G=S)
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Concurrent
Lines |
Point
of concurrence |
Name |
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Altitudes |
O |
Orthocenter |
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Angular bisectors |
I |
Incenter |
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Medians |
G |
Centroid(Center of gravity) |
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Perpendicular bisectors |
S/C |
Sircumcenter |
How to remember!!
OIGS stands for On India Government Services.
It is for All
(Altitudes)
Indian (Angular)
Middle(Medians)
class People
(Perpendicular
bisectors).
Note: In section 6.13 we
will be proving concurrency of these lines.
When we locate various points of concurrence in various
types of triangles we observe:
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Point
of concurrence |
Type of
triangle |
Location
of point of concurrence |
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Centroid (G) |
Any type of triangle |
Inside the triangle |
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In center(I) |
Any type of triangle |
Inside the triangle |
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Circumcentre(C,S) |
Acute angled triangle |
Inside the triangle |
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Right angled triangle |
Mid point of hypotenuse |
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Obtuse angled triangle |
Outside the triangle |
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Orthocenter(O) |
Acute angled triangle |
Inside the triangle |
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Right angled triangle |
Vertex of right angle |
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Obtuse angled triangle |
Outside the triangle |
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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? |
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6.5 Summary of learning
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No |
Points
to remember |
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1 |
The perpendicular drawn from the
vertex of a triangle to its opposite is called altitude. |
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2 |
Point of concurrency is a point
through which, three or more concurrent lines pass through |
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3 |
Orthocenter(O) is the point of
concurrency of the altitudes of a triangle |
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4 |
Centroid(G) is the point of
concurrency of the medians of a triangle |
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5 |
Circmcenter(S) is the point of
concurrency of the perpendicular bisectors of a triangle |
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6 |
Circumcircle is a circle which
passes through all the vertices of a triangle and its center is Circmcenter(S) |
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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. |
