Problem
GEO-B2-M04-P004 A Criterion for Spiral Similarity
#4
★★☆☆☆ Level 2 of 5
For a point \(P\), \(\frac{PA}{PB}=\frac{PC}{PD}\) and \(\angle APC=\angle BPD\). Prove that \(\triangle PAC \sim \triangle PBD\).
Use the similarity criterion with two proportional sides and the included angle.
In triangles \(PAC\) and \(PBD\), the sides adjacent to the angles at \(P\) are proportional: \(\frac{PA}{PB}=\frac{PC}{PD}\). The included angles are equal: \(\angle APC=\angle BPD\). Therefore \(\triangle PAC \sim \triangle PBD\). Hence \(P\) may be viewed as the centre of a spiral similarity sending \(AC\) to \(BD\).
This is a fundamental problem for the whole spiral similarity block.