$P$ is a point on side $B C$ of a parallelogram $A B C D$ . If $D P$ produced meets $A B$ produced at point $L$ , prove that :
(i) $D P : P L = D C : B L$ .
(ii) $D L : D P = A L : D C$ .

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Solution

(i) Consider $Δ D C P$ and $Δ B P L$
$∠ D P C = ∠ B P L$ [Vertically opposite angles]
$∠ D C P = ∠ P B L$ [Alternate angles as $D C ∥ A B$ ]
Therefore, $Δ D P C \sim Δ L P B$ by AA similarity rule.
Hence, $D P : P L = D C : B L$
(ii) Consider $Δ D L A$ and $P L B$
Since, $P B ∥ A D$ , therefore by Basic Proportionality Theorem,
$\frac{D L}{D P} = \frac{A L}{A B}$
Since, $A B = D C$
$∴ D L : D P = A L : D C$

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\frac{D P}{P L} = \frac{D C}{B L}

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In the adjoining figure ABCD is a parallelogram. ‘P’ is a point on BC, DP and AB are produced meet at L. Prove that DP : PL = DC : BL.

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