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CBSE NCERT Chapter 7 TRIANGLES
Example 3 and Theorem 7.2 and 7.3

Example 3 : Line-segment AB is parallel to another line-segment CD. O is the
mid-point of AD . Show that (i) ∆AOB ≅ ∆DOC (ii) O is also the
mid-point of BC.
Solution : (i) Consider ∆ AOB and ∆ DOC.
∠ ABO = ∠ DCO
(Alternate angles as AB || CD
and BC is the transversal)
∠ AOB = ∠ DOC
(Vertically opposite angles)
OA = OD (Given)
Therefore, ∆AOB ≅ ∆DOC (AAS rule)
(ii) OB = OC (CPCT)
So, O is the mid-point of BC.

Theorem 7.2 : Angles opposite to equal sides of an isosceles triangle are equal.
Proof : We are given an isosceles triangle ABC
in which AB = AC. We need to prove that
∠ B = ∠ C.
Let us draw the bisector of ∠ A and let D be
the point of intersection of this bisector of
∠ A and BC (see Fig. 7.25).
In ∆ BAD and ∆ CAD,
AB = AC (Given)
∠ BAD = ∠ CAD (By construction)
AD = AD (Common)
So, ∆ BAD ≅ ∆ CAD (By SAS rule)
So, ∠ ABD = ∠ ACD, since they are corresponding angles of congruent triangles.
So, ∠ B = ∠ C

Theorem 7.3 : The sides opposite to equal angles of a triangle are equal.
Given:- In triangle ABC ∠ B = ∠ C.
prove that :- AB = AC
Proof :
Let us draw the bisector of ∠ A and let D be the point of intersection of this bisector of ∠ A and BC .
In ∆ BAD and ∆ CAD,
∠ B = ∠ C. (Given)
∠ BAD = ∠ CAD (By construction) AD = AD (Common)
So, ∆ BAD ≅ ∆ CAD (By AAS rule)
So, AB = AC , since they are corresponding sides of congruent triangles.
So, AB = AC