In triangle ABC; angle ABC = 90° and P is a point on AC such that ∠PBC = ∠PCB. Show that : PA = PB.

In the right angled triangle ABC,

Let, ∠ACB = x

From figure,

⇒ ∠PCB = ∠ACB = x

⇒ ∠PBC = ∠PCB = x

By angle sum property of triangle,

⇒ ∠ABC + ∠ACB + ∠BAC = 180°

⇒ 90° + x + ∠BAC = 180°

⇒ ∠BAC = 180° - 90° - x = 90° - x

⇒ ∠BAP = ∠BAC = 90° - x.

From figure,

⇒ ∠ABP = ∠ABC - ∠PBC = 90° - x.

∴ ∠BAP = ∠ABP = 90° - x …..(1)

In △ BAP,

⇒ ∠BAP = ∠ABP [From (1)]

∴ PB = PA (Sides opposite to equal angles are equal)

Hence, proved that PA = PB.