Question

in a right triangle ABC a circle with a side AB as diameter is drawn to intersect the hypotenuse AC in P. prove that tangent to the circle at P bisects the sides BC.​

Answer 1

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Answer 2

Step-by-step explanation:

ABC 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 sides are equal)

(In a triangle, angles opposite to 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 ∆BPC

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

PBC + PCB = 180° - 2BPC = 180° - 90º = 90° ---- (2)

BPC = 90°

: BPQ + CPQ = 90° --- (3)

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

PBC + PCB = BPQ + CPQ

PCQ = CPQ (Sins, BPQ = PBQ)

Consider APQC,

PCQ = CPQ

:: PQ = QC ---- (4)

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

BQ = QC

Therefore, tangent at P bisects the side BC.

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