Question

In a right triangle ABC a circle is drawn with AB as the diameter which intersect hypotenuse AC at point P. Prove PB = PC.

Class X
Chapter: Circles

No spams!!​

Answer 1

Refer the attachment for the figure.

To prove : PB = PC

Construction :

Let PB be the Perpendicular bisector to AC.

Proof :

In ∆ ABC and ∆ BPC,

Angle B = Angle P ( Each 90° )

Angle C = Angle C ( Common angle )

By AA similarity,

∆ ABC ~ ∆ BPC... (1)

Similarly, We can prove

∆ ABC ~ ∆ APB..(2)

From (1) and (2),

∆ BPC ~ ∆ APB

Hence, PB = PC

Proved.

Answer 2

Hey..

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

Hope this helps u dude ✌☺️✌