Question

Measures of some angles in the figure are given. Prove that $\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{\mathrm{AQ}}{\mathrm{QC}}$

Answer 1

Given:
∠APQ = 60^∘
∠ABC = 60^∘
Since, the corresponding angles ∠APQ and ​∠APC are equal.
Hence, line PQ || BC.
$$\begin{gathered} \mathrm{In}△\mathrm{ABC},\mathrm{PQ}\parallel \mathrm{B}C \\ \frac{\mathrm{AP}}{\mathrm{PB}}=\frac{\mathrm{AQ}}{\mathrm{QC}}\left(\mathrm{By}\mathrm{Basic}\mathrm{proportionality}\mathrm{theorem}\right) \end{gathered}$$