Q. 94.5( 484 Votes )

In Fig. 10.13, XY and X′Y′ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X′Y′ at B. Prove that AOB = 90°.

Answer :

Let us join point O to C.


In ΔOPA and ΔOCA,


OP = OC (Radius of the same circle)


AP = AC (Tangents from point A)


AO = AO (Common side)


ΔOPA ΔOCA (SSS congruence criterion)

∠POA = COA … (i)

Similarly, ΔOQB ΔOCB

QOB = COB … (ii)


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


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


From equations (i) and (ii), it can be observed that 2COA + 2 COB = 180°

∠ COA + ∠ COB = 180° / 2

COA + COB = 90°

AOB = 90°

Hence Proved.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Quiz | Imp. Qs. on CirclesQuiz | Imp. Qs. on CirclesQuiz | Imp. Qs. on Circles37 mins
Quiz | Testing Your Knowledge on CirclesQuiz | Testing Your Knowledge on CirclesQuiz | Testing Your Knowledge on Circles32 mins
Short Cut Trick to Find Area of TriangleShort Cut Trick to Find Area of TriangleShort Cut Trick to Find Area of Triangle43 mins
Quiz | Areas Related to CirclesQuiz | Areas Related to CirclesQuiz | Areas Related to Circles43 mins
RD Sharma |  Area of Sector and SegmentsRD Sharma |  Area of Sector and SegmentsRD Sharma | Area of Sector and Segments25 mins
Quiz | Area Related with CirclesQuiz | Area Related with CirclesQuiz | Area Related with Circles47 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses