Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.5 AIME, harder Prove it

Example 5 As shown in Figure 5, let $P$ be a point inside $\square A B C D$ , and $\angle P B A=\angle P D A$ . Prove: $\angle P A B=\angle P C B$ .

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Solution

Prove that in Figure 5, draw $PE \parallel BC$ , and $PE = BC$ . It is easy to prove that quadrilateral $PEBC$ and quadrilateral $AEPD$ are both parallelograms.
$\begin{array}{l} \Rightarrow BE \parallel PC, AE \parallel DP \\ \Rightarrow \angle ABP = \angle ADP = \angle AEP \\ \Rightarrow A, E, B, P \text{ are concyclic} \\ \Rightarrow \angle PAB = \angle PEB = \angle PCB. \end{array}$

[Summary] By adding parallel lines to construct parallelograms, we use the property that opposite angles are equal to change the position of the "angle," and then complete the proof using the property of concyclic points.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.