Problem
GEO-B2-M04-P016 Hidden Equal Angles
#16
★★★★☆ Level 4 of 5
For a point \(P\), \(\frac{PA}{PB}=\frac{PC}{PD}\) and \(\angle APC=\angle BPD\). Prove that \(\angle PAC=\angle PBD\) and \(\frac{AC}{BD}=\frac{PA}{PB}\).
First prove the similarity \(\triangle PAC \sim \triangle PBD\).
Sides \(PA,PC\) and \(PB,PD\) are proportional, and the included angles are equal. Therefore \(\triangle PAC \sim \triangle PBD\). Hence corresponding angles are equal: \(\angle PAC=\angle PBD\). Also, from similarity, \(\frac{AC}{BD}=\frac{PA}{PB}\).
A standard problem on extracting consequences from a spiral similarity.