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

Given: 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 △ABC

∠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.