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  • If bisectors of opposite angles of a cyclic quadrilateral ABCD intersect the circle, circumscribing it at the points P and Q, prove that PQ is a diameter of the circle.

If bisectors of opposite angles of a cyclic quadrilateral ABCD intersect the circle, circumscribing it at the points P and Q, prove that PQ is a diameter of the circle.

Answers (1)

Given: Bisectors of opposite angles of a cyclic quadrilateral ABCD intersect the circle, circumscribing it at the points P and Q

Let ABCD be a cyclic quadrilateral

DP and QB are the bisectors of \angleADC and \angleABC respectively.

To Prove: PQ is the diameter of circle

Proof:

Joint QD and QC.

As, DP and QB are the bisectors of \angleADC and \angleABC respectively.

\angle 1 =\frac{1}{2}\angle CDA\; and \; \;\angle2 =\frac{1}{2} \angle CBA

Since ABCD is cyclic quadrilateral, sum of opposite angles is 180°

\angleCDA +\angleCBA = 180°

Divide both sides by 2,

\frac{1}{2} \angle CDA +\frac{1}{2} \angle CBA = 90^{\circ}

\angle 1 + \angle2 = 90°                                    … (i)

Now consider the segment QC. We know that angles in the same segment of a circle are equal. So,

\angle2 = \angle3

Putting the value in equation (i),

\angle1 + \angle3 = 90°

\anglePOQ = 90°

We know that angle in a semicircle is 90°

Hence PQ is a diameter of a circle.

Posted by

infoexpert24

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