Question

triangle ABC is a right angled triangle circle drawn with diameter AB and BC intersect at a point D shown that AB

=AD×AC​

Answer 1

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Step-by-step explanation:ΔABC is right angled triangle

∠ ABC = 90°

Drawn with AB as diameter that intersects AC at P, PQ is the tangent to the circle which intersects BC at Q.

Join BP.

PQ and BQ are tangents from an external point Q.

∴ PQ = BQ ---1 tangent from external point

⇒ ∠PBQ = ∠BPQ = 45° (isosceles triangle)

Given that, AB is the diameter of the circle.

∴ ∠APB = 90° (Angle subtended by diameter)

∠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.