$P$ 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) $D P : P L = D C : B L$

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Solution

Step $1$ : Drawing the figure

Given: $P$ is a point on sides $B C$ of a parallelogram $A B C D . D P$ produced meets $A B$ produced at point $L$ .

Step $2$ : Proving $D P : P L = D C : B L$

Given: $A B | | C D, A D | | B C$ as $A B C D$ is a parallelogram.
In $Δ D P C$ and $Δ L P B$ ,
$∠ D P C = ∠ L P B$ [Vertically opposite angles]
$∠ D C P = ∠ L B P$ [Alternate interior angles]
So, by $A A$ criterion, $Δ D P C$ ~ $Δ L P B$ .
$∴ \frac{D P}{L P} = \frac{P C}{P B} = \frac{C D}{B L}$ [ $∵$ the corresponding sides of similar triangles are proportional]
$\Rightarrow \frac{D P}{P L} = \frac{D C}{B L}$
$\Rightarrow D P : P L = D C : B L$

Hence, proved.

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Similar questions

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.

(i) \frac{DM}{MN} = \frac{DC}{BN} (ii) \frac{DN}{DM} = \frac{AN}{DC}

M is a point on the side BC of a parallelogram ABCD. DM when produced meets AB produced at N. Prove that

(i) $\frac{D M}{M N} = \frac{D C}{B N}$

(ii) $\frac{D N}{D M} = \frac{A N}{D C}$ .

P is a point on side BC of a parallelogram ABCD. If DP produced meets AB produced at point L, prove that:

I)DP:PL=DC:BL

ii)DL:DP =AL:DC

D P : P L = D C : B L

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