ABC is a triangle in which …

Given: ΔABC in which ∠B = 90°

Circle with diameter AB intersect the hypotenuse AC at P.

A tangent SPQ at P is drawn to meet BC at Q.

To prove: Q is mid point of BC.

Construction: Join PB.

Proof: SPQ is tangent and AP is chord at contact point P.

Therefore,∠2 = ∠3 [ since Angles in alternate segment of circle are equal]

∠2 = ∠1 [Vertically opposite angles]

∠3 = ∠1 …(i) [From above two relations]

∠ABC = 90° [Given]

OB is radius , therefore BC will be tangent at B.

Therefore,∠3 = 90° - ∠4 …(ii)

∠APB = 90° [∠ in a semi circle]

⇒ ∠C = 90° - ∠4....(iii)

From (ii) and (iii), ∠C = ∠3

Using (i), ∠C = ∠1

⇒ CQ = QP …(iv) [Sides opp. to = ∠s in ΔQPC]

∠4 = 90° - ∠3 [From fig.]

∠5 = 90° - ∠1

∠3 = ∠1

Therefore,∠4 = ∠5

⇒ PQ = BQ …(v) [Sides opp. to equal angles in ΔQPB]

From (iv) and (v),

BQ = CQ

Therefore, Q is mid-point of BC. Hence, proved.