Question

In a right triangle ABC in which∠B = 90º,a circle is drawn with ab as diameter intersecting the hypotenuse AC at P. Prove that the tangent to the circle at P bisect BC.

Answer 1

ΔABC is a right angled triangle.
∠ABC = 90°.
A circle is drawn with AB as diameter intersecting AC in P, PQ is the tangent to the circle which intersects BC at Q.

Join BP.

PQ and BQ are tangents drawn from an external point Q.

∴ PQ = BQ   -------------- (1)  (Length of tangents drawn from an external point to the circle are equal)

⇒ ∠PBQ = ∠BPQ    (In a triangle, angles opposit to equal sides are equal)

Given that, AB is the diameter of the circle.

∴ ∠APB = 90°  (Angle in a semi-circle is a right angle)

∠APB + ∠BPC = 180°   (Linear pair)

∴ ∠BPC = 180° – ∠APB = 180° – 90° = 90°

Consider ΔBPC,

∠BPC + ∠PBC + ∠PCB = 180°  (Angle sum property of a triangle)

∴ ∠PBC + ∠PCB = 180° – ∠BPC = 180° – 90° = 90°  ----------- (2)

∠BPC = 90°

∴ ∠BPQ + ∠CPQ = 90°    ...(3)

From equations (2) and (3), we get

∠PBC + ∠PCB = ∠BPQ + ∠CPQ

⇒ ∠PCQ = ∠CPQ  (Since, ∠BPQ = ∠PBQ)

Consider ΔPQC,

∠PCQ = ∠CPQ

∴ PQ = QC  ----------- (4)

From equations (1) and (4), we get

BQ = QC

Therefore, tangent at P bisects the side BC.

Hope This Helps :)