∆ABC and ∆DBC lie on the same side of BC, as shown in the figure.
Question:

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

 

Solution:

In $\triangle C A B, P Q \| A B$.

Applying Thales’ theorem, we get:

$\frac{C P}{P B}=\frac{C Q}{Q A} \quad \ldots(1)$

Similarly, applying Thales’ theorem in $\triangle B D C$, where $P R \| B D$, we get:

$\frac{C P}{P B}=\frac{C R}{R D} \quad \cdots(2)$

Hence, from $(1)$ and $(2)$, we have :

$\frac{C Q}{Q A}=\frac{C R}{R D}$

Applying the converse of Thales’ theorem, we conclude that $Q R \| A D$ in $\triangle A D C$.

This completes the proof.

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