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, 2009

Geometry Difficulty 6.0 AIME, harder Prove it JBMO

Problem:
A parallelogram $ABCD$ with obtuse angle $\angle ABC$ is given. After rotating the triangle $ACD$ around the vertex $C$ , we get a triangle $CD' A'$ , such that points $B$ , $C$ and $D'$ are collinear. The extension of the median of triangle $CD' A'$ that passes through $D'$ intersects the straight line $BD$ at point $P$ . Prove that $PC$ is the bisector of the angle $\angle BP D'$ .

Solution

Solution:
Let $AC \cap BD = \{X\}$ and $PD' \cap CA' = \{Y\}$ . Because $AX = CX$ and $CY = YA'$ , we deduce:
$\triangle ABC \cong \triangle CDA \cong \triangle CD' A' \Rightarrow \triangle ABX \cong \triangle CD' Y, \triangle BCX \cong \triangle D'A' Y$
It follows that
$\angle ABX = \angle CD' Y$
Let $M$ and $N$ be orthogonal projections of the point $C$ on the straight lines $PD'$ and $BP$ , respectively, and $Q$ is the orthogonal projection of the point $A$ on the straight line $BP$ . Because $CD' = AB$ , we have that $\triangle ABQ \cong \triangle CD' M$ .
We conclude that $CM = AQ$ . But, $AX = CX$ and $\triangle AQX \cong \triangle CNX$ . So, $CM = CN$ and $PC$ is the bisector of the angle $\angle BP D'$ .

Much shortened: $\triangle CD' Y \equiv \triangle CDX$ means their altitudes from $C$ are also equal, i.e. $CM = CN$ and the conclusion.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.