GEO-B2-M10-P024 Ratio of Diagonal Segments

Find two pairs of similar triangles: \(PAB\) and \(PDC\), then \(PAD\) and \(PBC\).

Since \(ABCD\) is cyclic, \(\angle PAB=\angle CAB=\angle CDB=\angle PDC\), and \(\angle PBA=\angle DBA=\angle DCA=\angle PCD\). Hence \(\triangle PAB\sim\triangle PDC\), so \(\frac{PB}{PC}=\frac{AB}{DC}\). Similarly, \(\triangle PAD\sim\triangle PBC\), hence \(\frac{PA}{PB}=\frac{AD}{BC}\). Multiplying these equalities gives \(\frac{PA}{PC}=\frac{AD}{BC}\cdot\frac{AB}{DC}=\frac{AB\cdot AD}{CB\cdot CD}\).

The final problem of the module: one result follows from two different similarities.