Question

prove that the intercept of a tangent between two parallel tangents to a circle subtends a right angle at the centre

Answer 1

Given: XY and X'Y' at are two parallel tangents to the circle wth centre O and AB is the tangent at the point C,which intersects XY at A and X'Y' at B.

To prove: ∠AOB = 90º .

Construction: Join OC.

Proof:
In ΔOPA and ΔOCA

OP = OC (Radii )

AP = AC (Tangents from point A)

AO = AO (Common )

ΔOPA ≅ ΔOCA (By SSS  criterion)

Therefore, ∠POA = ∠COA .... (1)   (By C.P.C.T)

Similarly , ΔOQB ≅ ΔOCB

∠QOB = ∠COB .......(2)

POQ is a diameter of the circle.

Hence, it is a straight line.

Therefore, ∠POA + ∠COA + ∠COB + ∠QOB = 180°

From equations (1) and (2), it can be observed that

2∠COA + 2∠COB = 180°

∴ ∠COA + ∠COB = 90°

∴ ∠AOB = 90°.

There is an alternate method too:

Consider CA and EB are the two tangents to the circle such that CA|| EB.

And AB intersect them at A and B respectively such that AB subtends angle AOB at the centre.

In ΔODA and ΔOCA

CA = AD (two tangemts from same side)

OA = OA (Common side)

OC = OD (Radius of circle)

⇒ ΔCAO ~ ΔDAO

So, ∠DOB = ∠DOA .............(1)

Similarly,

In ΔDOB and ΔBOEA

    ∠DOB = ∠EOB   ............(2)

Since COE is a diameter of the circle, it is a straight line.

Therefore,

∠COA + ∠DOA + ∠DOB + ∠EOB = 180° [From equations (1) and (2) ]

2∠DOA + 2∠DOB = 180°

⇒ ∠DOA + ∠DOB = 180°

⇒ ∠AOB = 90°

∴ the intercept of a tangent between two parallel tangents to a circle subtends a right angle at the centre.

Hope This Helps :)

Answer 2

Answer:

Step-by-step explanation: