Problem
GEO-B2-M04-P009 A New Segment from Spiral Similarity
#9
★★★☆☆ Level 3 of 5
For a point \(P\), it is known that \(\triangle PAC \sim \triangle PBD\). Prove that \(\frac{AC}{BD}=\frac{PA}{PB}\) and \(\angle ACP=\angle BDP\).
Write the correspondence of vertices in the given similarity.
In the similarity \(\triangle PAC \sim \triangle PBD\), the corresponding vertices are \(P \leftrightarrow P\), \(A \leftrightarrow B\), \(C \leftrightarrow D\). Therefore corresponding sides are proportional: \(\frac{AC}{BD}=\frac{PA}{PB}=\frac{PC}{PD}\). Corresponding angles are equal, so \(\angle ACP=\angle BDP\).
This problem reinforces reading correspondences in a spiral similarity.