Question

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

Answer 1

Given: XY and X'Y' are two parallel tangents to the circle with center O.
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 

AP = AC ......Tangents from point A

AO = AO .....Common

ΔOPA ≅ ΔOCA ...... By SSS  criterion

So, ∠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°.

Answer 2

Given: XY and X

′

Y

′

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=90

∘

Construction:Join OC

Proof:In △OPA and △OCA

OP=OC (radii)

AP=AC (Tangents from point A)

AO=AO (Common )

△OPA≅△OCA (By SSS criterion)

∴,∠POA=∠COA .... (1) (By C.P.C.T)

OQ=OC (radii)

BQ=BC (Tangents from point A)

BO=BO (Common )

△OQB≅△OCB

∠QOB=∠COB .......(2)

POQ is a diameter of the circle.

Hence, it is a straight line.

∴,∠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

∘