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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Solution

In $Δ$ DMC and $Δ$ MBN

$∠$ DMC = $∠$ NMB (vertically opposite angles)

$∠$ C = $∠$ B(alternate interior angles)

now we can say triangles are similar

therefore

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

Similarly,

In $Δ$ DNA and $Δ$ DMC

$∠$ A = $∠$ C(opp. angles of the parallelogram)

$∠$ N = $∠$ D (alternate interior angles)

hence we can say triangles are similar

$\frac{D N}{D M} = \frac{A N}{D C}$ (by CPCT)

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Similar questions

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

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

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

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

D P : P L = D C : B L

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