Prove that the median drawn from the vertex P of an isosceles triangle △PQR with PQ = PR is perpendicular to QR and bisects ∠P.

Let PQR be an isosceles triangle with PQ = PR.

Draw a median from vertex P to the side QR. Let this median be PS, where S is the midpoint of QR.

In △PQS and △PRS,

⇒ PQ = PR (△PQR is an isosceles triangle).

⇒ QS = RS (Since PS is the median, S is the midpoint of QR).

⇒ PS = PS (Common side to both triangles).

By the SSS (Side-Side-Side) congruence criterion,

△PQS ≅ △PRS

We know that,

Corresponding parts of congruent triangles are congruent.

⇒ ∠PSQ = ∠PSR = x (let)

From figure,

⇒ ∠PSQ + ∠PSR = 180° [Linear pair]

⇒ x + x = 180°

⇒ 2x = 180°

⇒ x = $\dfrac{180°}{2}$ = 90°.

⇒ ∠PSQ = ∠PSR = 90°.

Thus, PS is perpendicular to QR.

So, PS ⊥ QR.

⇒ ∠QPS = ∠RPS (By C.P.C.T.C.)

Thus, PS bisects ∠P.

Hence, the median drawn from the vertex P of an isosceles triangle △PQR is perpendicular to QR and bisects ∠P.