Question

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

Answer 1

According to the question,

In a right angle ΔABC is which ∠B = 90°, a circle is drawn with AB as diameter intersecting the hypotenuse AC at P.

Also PQ is a tangent at P

To Prove: PQ bisects BC i.e. BQ = QC

Proof: ∠APB = 90°  ...[Angle in a semicircle is a right-angle]

∠BPC = 90° ...[Linear Pair]

∠3 + ∠4 = 90° ...[1]

Now, ∠ABC = 90°

So in ΔABC

∠ABC + ∠BAC + ∠ACB = 180°

90° + ∠1 + ∠5 = 180°

∠1 + ∠5 = 90°  ...[2]

Now, ∠1 = ∠3  ...[Angle between tangent and the chord equals angle made by the chord in alternate segment]

Using this in [2] we have

∠3 + ∠5 = 90°  ...[3]

From [1] and [3] we have

∠3 + ∠4 = ∠3 + ∠5

∠4 = ∠5

QC = PQ  ...[Sides opposite to equal angles are equal]

But also, PQ = BQ  ...[Tangents drawn from an external point to a circle are equal]

So, BQ = QC

i.e. PQ bisects BC.