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


Hence proved

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

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