To prove two angles are equal when some angles are supplementary in a parallelogram

Question

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

Answer 1

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

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.