Question

Two circles intersect at P and Q Through P two straight lines APB and CQD are drawn to meet the circles at AB.C and D AC and BD when produced meet at O Show that OAQB is a cyclic quadrilateral​

Answer 1

Refer below answer for the question

Answer 2

OAQB is a cyclic quadrilateral​

Step-by-step explanation:

Show that : OAQB is a cyclic quadrilateral

i.e,

 $\angle O + \angle AQB = 180$°

construction : join PQ

proof:

$\angle PQB =\angle PDB$...(angle in the same segment)..(1)

$\angle ACP +\angle AQP = 180$..(AQPC is a cyclic quadrilateral)..(2)

but,

$\angle ACP = \angle ACD = \angle O +\angle CDB$..(sum of the interior angles)

now, replacing angle ACP from eq (2)

we get,

$180- \angle AQP = \angle O +\angle CDB$

using eq(1)

$180= \angle AQP +\angle O +\angle PQB$

$180= \angle O +\angle AQB$

hence proved

OAQB is a cyclic quadrilateral​

#Learn more:

AB is a diameter of a circle O Chord CD is equal to radius OC. If AC and BD when produced intersect at P, prove that APB is 60°

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