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.