In this chapter we will learn about the concept of “Congruence” of triangles.
In simple, any two geometrical figures, exactly of the same size and shape are congruent.
Congruence of a line segment: Two line segment and
are congruent if and only if their lengths are equal. i.e. if
Congruence of angles: Two angles are congruent if and only if their measures are equal. i.e.
Congruence of triangles:
Consider two and
(as shown in the figure)
Two triangles are congruent if and only if one of them can be made to superpose on the other so as to cover it exactly.
Two triangles are congruent if and only if there exists a correspondence between their vertices such that the corresponding sides and the corresponding angles of the two triangles are equal. This means that if:
and
and
Also if and
Theorems of Congruence
Theorem 1: S.A.S (Side Angle Side): Two triangles are congruent if two sides and the included angle of one triangle are equal to the corresponding sides and the included angle of the other triangle.
Example 1: In the adjoining figure, is the mid point of
and
. Prove that (i)
(ii)
(iii)
Solution: Consider and
(Given)
(Given)
(vertically opposite angle)
Therefore (By S.A.S theorem)
Since the triangles are congruent, (corresponding sides of congruent triangles are equal)
Also since the two triangles are congruent, (corresponding angles are also equal). Hence
(alternate angles are equal)
Example 2: In the adjoining figure, it is given that
and
. Prove that
.
Solution: is common segment of
and
and
Therefore
Now consider and
(given)
(Proved above)
is common
Therefore
Theorem 2: A.S.A (Angle Side Angle): Two triangles are congruent if two angles and the included side of one triangle are equal to the corresponding two angles and the included side of the other triangle.
Example 3: In the adjoining triangle, diagonal
is a quadrilateral
bisects the
and
. Prove that
and
.
Solution:
Given: and
Consider and
(given)
(given)
and is common.
Therefore (A.S.A theorem)
Hence we can say that and
Example 4:
is a line segment,
and
are two equal line segments draw on opposite sides of line
such that
. If
and
intersect each other at
, prove that (i)
(ii)
and
bisect each other
Solution:
Given that and
is the transversal.
Therefore
Similarly, and
is the transversal.
Therefore
Now consider and
(given)
Therefore
and
Theorem 3.1 A.A.S (Angle Angle Side): If any two angles and a not included side of one triangle are equal to the corresponding angles and side of another triangle, then the two triangles are congruent.
Solution / Proof:
In the given and
, we have
,
and
.
Therefore
Hence we see that with and
Hence (A.S.A Theorem).
Theorem 3.2: If two angles of a triangle are equal, then sides opposite to them are also equal.
Solution:
Given:
Draw the bisector of . Now consider
and
We have (angle bisector)
and is common or equal.
Hence by (A.A.S theorem)
Theorem 3.3: If the altitude of one vertex of a triangle bisects the opposite side, then the triangle is isosceles.
Solution:
In the is the altitude
(given)
and is common.
Therefore
Theorem 3.4: In an isosceles triangle altitude from the vertex bisects the base.
Solution:
In , given
(Altitude)
And is common.
Therefore (A.A.S theorem)
Theorem 3.5: If the bisector of the vertical angle of a triangle bisects the base of the triangle, then the triangle is isosceles.
Solution:
Extend to
such that
Given: and
Now, consider and
Therefore (given) ,
(constructed)
and (vertically opposite angles)
Hence (S.A.S theorem)
Therefore and
we know,
. Hence
is an isosceles triangle.
Theorem 4: S.S.S (Side Side Side): Two triangles are congruent if the three sides of one triangle are equal to the corresponding three sides of the other triangle.
Given two and
such that
and
To Prove:
Solution:
In , draw
such that
and
.
Join and
Consider, and
We have
(given)
(by construction)
And (by construction)
Therefore (by S.A.S theorem)
and
Since and
In (angle opposite to the equal sides are equal)
Similarly, in In (angle opposite to the equal sides are equal)
Hence
Now consider and
(given)
(just proved above)
(given)
Therefore (by S.A.S theorem). Hence proven.
Theorem 5: R.H.S (Right Angle Hypotenuse Side): Two right triangles are congruent if the hypotenuse and one side of one triangle are respectively equal to the hypotenuse and the one side of the other triangle.
Given two right angled and
.
and
Also given, and
To Prove:
Solution:
Consider and
(by construction)
(by construction)
(given)
Therefore (By S.A.S theorem)
and
(angle opposite to the equal sides are equal)
Therefore
In and
, we have
Now consider and
(given)
(proved above)
(given)
Therefore (by S.A.S theorem)
Inequality Relations in a Triangle
Theorem 1: If two sides of a triangle are unequal, the longer side has greater angle opposite to it.
Given in
Find a point on
such that
… … … (i)
Consider
We know (exterior angle is the sum of other two angles)
or
… … … … … (ii)
Therefore from (i) and (ii) we get
and
… … … … … (iii)
From we get
… … … … … (iv)
From (iii) and (iv) we get
. Hence proven.
Theorem 2: In a triangle the greater angle has the longer side opposite to it (converse of Theorem 1).
In , given
To Prove:
Solution:
There are three possible cases (i) (ii)
(iii)
Case 1: If , then
(angles opposite to equal sides are equal). But that is not true as it is given that
Case 2: When (longer side has the greater angle opposite it). This not true either.
Case 3: Hence is the only possibility that is left.
Theorem 3: The sum of any two sides of a triangle is greater than the third side.
Given
To Prove: and
Construction: Extend to
such that
Join
Solution:
In
(by construction)
(angles opposite to equal sides are equal)
(since side opposite to the greater angle is larger)
(since
by construction)
Similarly, you can prove that and
Theorem 4: Of all the line segment that can be drawn to a given line, from a point, not lying on it, the perpendicular line segment is the shortest.
Given that is a straight line and point
is not lying on the line.
and
is any other point
.
Prove:
Solution: In , we have
(sum of the angle of a triangle is
)
. Hence proven.
Pythagoras Theorem
Theorem 1 (Pythagoras Theorem): In a a right triangle, the square of the hypotenuse is equal to the sum of the square of the two sides.. If in a , if
is the hypotenuse, then
.
Construction: Extend side to point
such that
. At
draw
and cut off
. Join
.
Solution:
Consider and
(by construction)
(right angles)
(by construction)
(by S.A.S theorem)
and
We know
Also
That is is a right angle triangle.
Therefore at point we have
is a trapezium
Now, area of trapezium = area of
+ area of
+ area of
Theorem 2 (Converse of the Pythagoras Theorem): In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite to this side is a right angle.
Given, such that
To Prove:
Construction: Draw another such that
and
Solution:
Since by construction, is a right angled triangle, by Pythagoras theorem,
In and
, we have
and
Therefore
Hence is a right angled triangle.