Question 133

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:

Solution

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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