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The point 'P' is situated inside the parallelogram ABCD such that the angles APB and CPD are supplementary.Prove that angles PBC and PDC are equal

thanasissdr
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lokesh sangabattula
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1 Answers1

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According to the given, $\alpha + \beta = 180^0$.

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Construction:

1) Through P, draw PM // BC. 2) Join PB. 3) Through C, draw CM // PB. 4) Join DM.

Then, PBCM is // gm. Hence, PM = BC.

In the quadrilateral ADMP and //gm ABCD, from AD = BC = PM and AD // BC // PM, we can say that ADMP is a // gm.

This means ∠APB is successfully translated to ∠DMC. Therefore, $\theta = \alpha$.

$\theta + \beta = \alpha + \beta = 180^0$ implies PDMC is a cyclic quadrilateral.

By angles in the same segment, $\gamma = y$.

Also, by properties of //gm, $y = \delta$.

Result follows.

Mick
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  • What your construction also shows: if P inside a # lies s.t. $\angle CBP = \angle PDC$ then also $\angle BAP = \angle PCB$ (and vice versa). Surprising! – coproc Jul 20 '15 at 16:16
  • @coproc Yes. One is an instance of the other. – Mick Jul 20 '15 at 16:25