# M is a point on the side BC of a parallelogram ABCD.

Question:

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

(i) $\frac{\mathrm{DM}}{\mathrm{MN}}=\frac{\mathrm{DC}}{\mathrm{BN}}$

(ii) $\frac{\mathrm{DN}}{\mathrm{DM}}=\frac{\mathrm{AN}}{\mathrm{DC}}$

Solution:

Given: ABCD is a parallelogram
To prove:

(i) $\frac{\mathrm{DM}}{\mathrm{MN}}=\frac{\mathrm{DC}}{\mathrm{BN}}$

(ii) $\frac{\mathrm{DN}}{\mathrm{DM}}=\frac{\mathrm{AN}}{\mathrm{DC}}$

Proof: In $\triangle D M C$ and $\triangle N M B$

$\angle \mathrm{DMC}=\angle \mathrm{NMB} \quad$ (Vertically opposite angle)

$\angle \mathrm{DCM}=\angle \mathrm{NBM} \quad$ (Alternate angles)

By AAA- similarity

$\triangle \mathrm{DMC} \sim \triangle \mathrm{NMB}$

$\therefore \frac{\mathrm{DM}}{\mathrm{MN}}=\frac{\mathrm{DC}}{\mathrm{BN}}$

Now, $\frac{\mathrm{MN}}{\mathrm{DM}}=\frac{\mathrm{BN}}{\mathrm{DC}}$

Adding 1 to both sides, we get

$\frac{\mathrm{MN}}{\mathrm{DM}}+1=\frac{\mathrm{BN}}{\mathrm{DC}}+1$

$\Rightarrow \frac{\mathrm{MN}+\mathrm{DM}}{\mathrm{DM}}=\frac{\mathrm{BN}+\mathrm{DC}}{\mathrm{DC}}$

$\Rightarrow \frac{\mathrm{MN}+\mathrm{DM}}{\mathrm{DM}}=\frac{\mathrm{BN}+\mathrm{AB}}{\mathrm{DC}} \quad[\because \mathrm{ABCD}$ is a parallelogram $]$

$\Rightarrow \frac{\mathrm{DN}}{\mathrm{DM}}=\frac{\mathrm{AN}}{\mathrm{DC}}$