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

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Solution

Given: ABCD is a parallelogram. DP produced meets AB produced at L
In $△ L A D$ ,
BP II AD (Given ABCD is a parallelogram and P lies on BC)
$∴$ , $\frac{D L}{D P} = \frac{A L}{A B}$ (Basic proportionality theorem)
or $\frac{D L}{D P} = \frac{A L}{D C}$ (AB = DC opposite sides of parallelogram)

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Similar questions

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 L : D P = A L : D C

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

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

△ B E C ≅△ D C F

D P : P L = D C : B L

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