Question

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

$$\begin{gathered} \left(\mathrm{i}\right)\frac{\mathrm{DM}}{\mathrm{MN}}=\frac{\mathrm{DC}}{\mathrm{BN}} \\ \left(\mathrm{ii}\right)\frac{\mathrm{DN}}{\mathrm{DM}}=\frac{\mathrm{AN}}{\mathrm{DC}} \end{gathered}$$

Answer 1

Given: ABCD is a parallelogram
To prove:
$$\begin{gathered} \left(\mathrm{i}\right)\frac{\mathrm{DM}}{\mathrm{MN}}=\frac{\mathrm{DC}}{\mathrm{BN}} \\ \left(\mathrm{ii}\right)\frac{\mathrm{DN}}{\mathrm{DM}}=\frac{\mathrm{AN}}{\mathrm{DC}} \end{gathered}$$
Proof: In △DMC and △NMB
$\angle$DMC =$\angle$NMB (Vertically opposite angle)
$\angle$DCM =$\angle$NBM (Alternate angles)
By AAA- similarity
△DMC ~ △NMB
$\therefore \frac{\mathrm{DM}}{\mathrm{MN}}=\frac{\mathrm{DC}}{\mathrm{BN}}$
$\mathrm{Now},\frac{\mathrm{MN}}{\mathrm{DM}}=\frac{\mathrm{BN}}{\mathrm{DC}}$
Adding 1 to both sides, we get
$$\begin{gathered} \frac{\mathrm{MN}}{\mathrm{DM}}+1=\frac{\mathrm{BN}}{\mathrm{DC}}+1 \\ \Rightarrow \frac{\mathrm{MN}+\mathrm{DM}}{\mathrm{DM}}=\frac{\mathrm{BN}+\mathrm{DC}}{\mathrm{DC}} \\ \Rightarrow \frac{\mathrm{MN}+\mathrm{DM}}{\mathrm{DM}}=\frac{\mathrm{BN}+\mathrm{AB}}{\mathrm{DC}}\left[\because \mathrm{ABCD}\mathrm{is}\mathrm{a}\mathrm{parallelogram}\right] \\ \Rightarrow \frac{\mathrm{DN}}{\mathrm{DM}}=\frac{\mathrm{AN}}{\mathrm{DC}} \end{gathered}$$