Question

∆ABC and ∆DBC lie on the same side of BC, as shown in the figure. From a point P on BC, PQ ∥ AB and PR ∥ BD are drawn, meeting AC at Q and CD at R, respectively. Prove that QR ∥ AD.

Answer 1

$$\begin{gathered} \mathrm{In}△CAB,PQ\parallel AB. \\ \mathrm{Applying}\mathrm{Thales}\text{'}\mathrm{theorem},\mathrm{we}\mathrm{get}: \\ \frac{CP}{PB}=\frac{CQ}{QA}...\left(1\right) \end{gathered}$$
Similarly, applying Thales' theorem in $△BDC,\mathrm{where}PR\parallel BD$, we get:
$$\begin{gathered} \frac{CP}{PB}=\frac{CR}{RD}...\left(2\right) \\ \mathrm{Hence},\mathrm{from}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{have}: \\ \frac{CQ}{QA}=\frac{CR}{RD} \end{gathered}$$

Applying the converse of Thales' theorem, we conclude that $QR\parallel ADin△ADC$.
This completes the proof.