Answer :

Let the circle with centre O and chord PQ with tangents from point A as AP and AQ as shown:


We have to prove that APQ = AQP


Consider ΔOPQ


OP = OQ …radius


Hence ΔOPQ is an isosceles triangle


OPQ = OQP …base angles of isosceles triangle …(a)


As radius OP is perpendicular to tangent AP at point of contact P


APO = 90°


From figure APO = APQ + OPQ


90° = APQ + OPQ


APQ = 90° - OPQ …(i)


As radius OQ is perpendicular to tangent AQ at point of contact Q


AQO = 90°


From figure AQO = APQ + OPQ


90° = AQP + OQP


AQP = 90° - OQP


Using (a)


AQP = 90° - OPQ …(ii)


Using (i) and (ii), we can say that


APQ = AQP


Hence proved


Hence, the tangents at the extremities of any chord of a circle make equal angles with the chord


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
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