In the adjoining figure, △ABC is right-angled at B and P is the mid-point of AC. Show that, PA = PB = PC.

Given,

P is mid-point of AC,

∴ PA = PC ………(1)

From figure,

PQ // CB

∠AQP = ∠ABC = 90° (Corresponding angles are equal)

⇒ ∠AQP + ∠PQB = 180° (Linear pair)

⇒ 90° + ∠PQB = 180°

⇒ ∠PQB = 180° - 90°

⇒ ∠PQB = 90°

By converse of mid-point theorem,

A line drawn through the midpoint of one side of a triangle, and parallel to another side, will bisect the third side.

In △ABC,

Since, P is the mid-point of AC and PQ // CB, thus by converse of mid-point theorem,

PQ will bisect AB, thus Q is the mid-point of AB.

In △AQP and △BQP,

⇒ ∠AQP = ∠PQB (Each equal to 90°)

⇒ PQ = PQ (Common side)

⇒ AQ = BQ (Q is the mid-point of AB)

Thus, △AQP ≅ △BQP. (By S.A.S. axiom)

∴ PB = PA (Corresponding parts of congruent triangles are equal) ………(2)

From (1) and (2), we get :

⇒ PB = PA = PC

Hence, proved that PA = PB = PC.