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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