Let $ABCD$ be a parallelogram. Suppose that there exists a point $P$ in the interior of the parallelogram which is on the perpendicular bisector of $AB$ and such that $\angle PBA = \angle ADP$ Show that $\angle CPD = 2 \angle BAP$
Problem
Source: Swiss TST 2015. Problem 10
Tags: geometry, parallelogram
RC.
30.07.2017 09:12
Bisector of \(AB\) what does this mean?
MeineMeinung
30.07.2017 11:34
I assume that you mean perpendicular bisector of $AB$.
Let $PA$ intersects $BC$ at $A'$.
Notice that $\angle DPA = 180^{\circ} - \angle PAD - \angle ADP = 180^{\circ} - (\angle PAD + \angle PAB ) = 180^{\circ} - \angle{DAB} = \angle ADC = \angle ABA'$.
It is also clear that $\angle A'AB = \angle ADP$, so $\triangle AA'B \sim \triangle DAP \implies \frac{PA}{PD} = \frac{BA'}{BA} \implies \frac{BP}{DP} = \frac{BA'}{DC}$.
But since $\angle{A'BP} = \angle A'BA - \angle PBA = \angle ADC - \angle PDA = \angle CDP$, we are able to conclude that $\triangle A'BP \sim CDP$, so $\angle CPD = \angle A'PB = 2\angle BAP$.
RC.
30.07.2017 12:16
tchebytchev wrote: Let $ABCD$ be a parallelogram. Suppose that there exists a point $P$ in the interior of the parallelogram which is on the bisector of $AB$ and such that $\angle PBA = \angle ADP$ Show that $\angle CPD = 2 \angle BAP$ All what you have to do here is draw \(\Delta CP'D \cong \Delta BPA\) such that \(CD\) seperates \(P\) and \(P'\) . Now, \(\angle DAC + ADP' = 180^{\circ} => AP || DP'\) and \(AP = DP\) \(=> ADP'P\) is a parallelogram. \(=> DA || PP' => \angle DPP' = \angle ADP = \angle DCP' => PDP'C\) is cyclic in nature. \(=>\angle CPD = 180 - \angle DP'C = 2 \times \angle BAP\)