Question

prove that the interception
of the tangent between the two parallel tangents suspended at the 90°​

Answer 1

Answer:

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.

Step-by-step explanation: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°.

Answer 2

Given : Line l is a tangent to the circle with centre O at the point of contact A.

To prove : line l ^ radius OA.

Proof : Assume that, line l is not

perpendicular to seg OA.

Suppose, seg OB is drawn

perpendicular to line l.

Of course B is not same as A.

Now take a point C on line l

such that A-B-C and

BA = BC .

Now in, D OBC and D OBA

seg BC @ seg BA ........ (construction)

Ð OBC @ Ð OBA ....... (each right angle)

seg OB @ seg OB

\ D OBC @ D OBA .......... (SAS test)

\ OC = OA

But seg OA is a radius.

\seg OC must also be radius.

\ C lies on the circle.

That means line l intersects the circle in two

distinct points A and C.

But line l is a tangent. ........... (given)

\ it intersects the circle in only one point.

O A

l

Fig. 3.10

Fig. 3.11

A

B

C

O

l

Our assumption that line l is not perpendicular to radius OA is wrong.

\ line l ^ radius OA

hope this helps you...

plz mark us brilliant....