Question

In the adjoining figure P and Q are two points on equal sides AB and AC of an isosceles triangle ABC such that AP=AQ, prove that BQ = CP

Answer 1

Hint: In this question, first we have to find out the difference between AB, AP, and AC, AQ by using the property of the Isosceles triangle. Then congruence the triangles ∆PBC & ∆QCB by using S-A-S (Side-Angle-Side property) to reach the final solution. Also, we know that in an isosceles triangle, two sides are of equal length and corresponding base angles are equal.

Complete step by step answer:
Consider the isosceles ∆ABC
In ∆ABC, we have
$AB = AC$ (given)................ (1)
$AP = AQ$ (Given)................... (2)
If we consider the difference as
$AB – AP$ & $AC – AQ$
$\Rightarrow BP = CQ$ from given diagram………………. (3)

Consider the triangles ∆PBC & ∆QCB
In ∆PBC & ∆QCB,
$BP = CQ$ (From 3)
$\angle PBC = \angle QCB$ (Base angles of isosceles triangle are congruent
$\Rightarrow BC = CB$ (Common Sides)
By S-A-S postulate
$\Rightarrow $∆PBC ≅ ∆QCB
We have corresponding sides of congruent triangles are equal.

$\therefore$ CP = BQ. Hence proved.

Note:
Whenever we face such types of questions, use Side-Angle-Side(S-A-S) postulate to prove congruence of two triangles. Then use corresponding sides and angles of congruent triangles, we will get the required proof.