Q. 125.0( 4 Votes )

Show that the tangents at the extremities of any chord of a circle make equal angles with the chord.

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


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