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AB is a diameter of a circle having centre at O, PQ is a chord which does not intersect AB. Join AP and BQ. If $$\ \angle$$BAP=$$\angle$$ABQ, then ABQP is a:
Given, AB is a diameter of circle
$$\ \angle$$BAP = $$\angle$$ABQ and
P, Q are points on the circle so PQ should be parallel to AB as shown in figure
$$\therefore\ $$ PQ||AB and $$\ \angle$$BAP = $$\angle$$ABQ
$$=$$> ABQP is a Cyclic trapezium
Hence, option A is correct answer
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