Mathematics
In the given figure APB and CQD are two straight lines, then :
AB || CD
AC || PQ
PQ || BD
AC || BD
Circles
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Answer
Let ∠BPQ = x and ∠DQP = y
We know that,
The exterior angle of a cyclic quadrilateral is equal to the opposite interior angle.
From figure,
In cyclic quadrilateral APQC,
∠A = ∠DQP = y and ∠C = ∠BPQ = x
From figure,
APB is a straight line.
∴ ∠APQ + ∠BPQ = 180°
⇒ ∠APQ + x = 180°
⇒ ∠APQ = 180° - x
CQD is a straight line.
∴ ∠CQP + ∠DQP = 180°
⇒ ∠CQP + y = 180°
⇒ ∠CQP = 180° - y
In cyclic quadrilateral PQDB,
∠B = ∠CQP = 180° - y and ∠D = ∠APQ = 180° - x
⇒ ∠A + ∠B = y + (180° - y) = 180°
⇒ ∠C + ∠D = x + (180° - x) = 180°
We know that,
Sum of adjacent angles in a trapezium is 180°.
∴ ABDC is a trapezium.
∴ AC || BD.
Hence, Option 4 is the correct option.
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