Question

Let ABC be a triangle and D and E be two points on side AB such that AD = BE. If DP $\parallel$ BC and EQ $\parallel$ AC, then prove that PQ $\parallel$ AB.

Answer 1

In $△$ABC, we have

DP || BC and EQ || AC

$\therefore$ $\frac{AD}{DB}$ = $\frac{AP}{PC}$ and $\frac{BE}{EA}$ = $\frac{BQ}{QC}$

$\Rightarrow$ $\frac{AD}{DB}$ = $\frac{AP}{PC}$ and $\frac{AD}{DB}$ = $\frac{BQ}{QC}$

$$\left[\begin{array}{c}EA-ED+DA-ED+BE-BD\\ (\because AD-BE)\end{array}\right]$$

$\Rightarrow$ $\frac{AP}{PC}$ = $\frac{BQ}{QC}$

$\Rightarrow$ In a $△$ABC, P and Q divide sides CA and CB respoectively in the same ratio.

$\Rightarrow$ PQ || AB