ABCD is a parallelogram, P is a point on side BC such that DP produced meets AB produced at L.. Prove that

(i) $\frac{D P}{P L} = \frac{D C}{B L}$

(ii) $\frac{D L}{D P} = \frac{A L}{D C}$

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Solution

Given A parallelogram ABCD in which P is a point on side BC such that DP produced meets AB produced at L.
PROOF (i) In $△ A L D$ , we have
$B P | | A D$

$∴$ $\frac{L B}{B A} = \frac{L P}{P D}$

$\Rightarrow$ $\frac{B L}{A B} = \frac{P L}{D P}$

$\Rightarrow$ $\frac{B L}{D C} = \frac{P L}{D P}$ [because AB=DC]

$\Rightarrow$ $\frac{D P}{P L} = \frac{D C}{B L}$ [Taking reciprocals of both sides] $[H e n c e p r o v e d]$

(ii) From (i), we have

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

$\Rightarrow$ $\frac{P L}{D P} = \frac{B L}{D C}$ [Taking reciprocals of both sides]

$\Rightarrow$ $\frac{P L}{D P} = \frac{B L}{A B}$ [because DC=AB]

$\Rightarrow$ $\frac{P L}{D P} + 1 = \frac{B L}{A B} + 1$ [Adding 1 on both sides]

$\Rightarrow$ $\frac{D P + P L}{D P} = \frac{B L + A B}{A B}$

$\Rightarrow$ $\frac{D L}{D P} = \frac{A L}{A B}$

$\Rightarrow$ $\frac{D L}{D P} = \frac{A L}{D C}$ $[∵ A B = D C]$ $[H e n c e p r o v e d]$

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Similar questions

$P$ 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) $D L : D P = A L : D C$ .

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.

Q. P is a point on side BC of a parallelogram ABCD. If DP produced meet AB produced at point L, then

D P : P L = D C : B L

Q. P is a point on side BC of a parallelogram ABCD. If DP produced meets AB produced at point L, then

DL : DP = AL : BC.
If the above statement is true then mention answer as 1, else mention 0 if false

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

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ABCD is a parallelogram, P is a point on side BC such that DP produced meets AB produced at L.. Prove that(i) DPPL=DCBL(ii) DLDP=ALDC

ABCD is a parallelogram, P is a point on side BC such that DP produced meets AB produced at L.. Prove that

(i) $\frac{D P}{P L} = \frac{D C}{B L}$

(ii) $\frac{D L}{D P} = \frac{A L}{D C}$