Question

Use the given figure to prove that, AB = AC.

Answer 1

In ΔAPQ,
AP = AQ .........[Given]
∴ ∠APQ = ∠AQP .........(i) [angles opposite to equal sides are equal]

In ΔABP,
∠APQ = ∠BAP + ∠ABP   .......(ii) [Ext. the angle is equal to the sum of opp. int. angles]

In ΔAQC,
∠AQP = ∠CAQ + ∠ACQ   ...(iii) [Ext. angle is equal to sum of opp. int. angles]
From (i), (ii) and (iii)
∠BAP + ∠ABP = ∠CAQ + ∠ACQ
But, ∠BAP = ∠CAQ .......[Given]
⇒ ∠CAQ + ∠ABP = ∠CAQ + ∠ACQ
⇒ ∠ABP = ∠CAQ + ∠ACQ − ∠CAQ
⇒ ∠ABP = ∠ACQ
⇒ ∠B = ∠C ...........(iv)

In ΔABC,
∠B = ∠C
⇒ AB = AC ......[Sides opposite to equal angles are equal]