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

Formula Used: Tangents drawn from an external point to a circle are equal.

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,

We know that angle between tangent and the chord equals angle made by the chord in alternate segment.

1 = 3

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

So, BQ = QC

i.e. PQ bisects BC.