Question
$\triangle\text{ABC}$ and $\triangle\text{DBC}$ 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
In $\triangle\text{CAB},\text{PQ }||\text{ AB}.$
Applying Thales' theorem, we get:
$\frac{\text{CP}}{\text{PB}}=\frac{\text{CQ}}{\text{QA}}\dots(1)$
Similarly, applying Thales' theorem $\triangle\text{BDC},$ where $PR || BD$, we get:
$\frac{\text{CP}}{\text{PB}}=\frac{\text{CR}}{\text{RD}}\dots(2)$
Hence, from $(1)$ and $(2),$ we have:
$\frac{\text{CQ}}{\text{QA}}=\frac{\text{CR}}{\text{RD}}$
Applying the converse of Thales' theorem, we conclude that $QR || AD$ in $\triangle\text{ADC}.$
This completes the proof.
Applying Thales' theorem, we get:
$\frac{\text{CP}}{\text{PB}}=\frac{\text{CQ}}{\text{QA}}\dots(1)$
Similarly, applying Thales' theorem $\triangle\text{BDC},$ where $PR || BD$, we get:
$\frac{\text{CP}}{\text{PB}}=\frac{\text{CR}}{\text{RD}}\dots(2)$
Hence, from $(1)$ and $(2),$ we have:
$\frac{\text{CQ}}{\text{QA}}=\frac{\text{CR}}{\text{RD}}$
Applying the converse of Thales' theorem, we conclude that $QR || AD$ in $\triangle\text{ADC}.$
This completes the proof.
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