Question : AB is the diameter of a circle having a centre at O. PQ is a chord that does not intersect AB. Join AP and BQ. If $\angle$BAP = $\angle$ABQ, then ABQP is a:
Option 1: cyclic square
Option 2: cyclic trapezium
Option 3: cyclic rhombus
Option 4: cyclic rectangle
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Correct Answer: cyclic trapezium
Solution :
$\angle$BAP = $\angle$ABQ and AB $\neq$ PQ
Now, $\angle$BAP + $\angle$BQP = 180°
⇒ $\angle$ABQ + $\angle$BQP = 180°
⇒ AB || PQ
So, ABQP is a cyclic trapezium.
Hence, the correct answer is cyclic trapezium.
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