Question

M' is a point on the side BC of a parallelogram ABCD. DM when produced meets AB produced at N. Prove that DM /MN= DC/BN

Answer 1

Proved below.

Step-by-step explanation:

Given:

ABCD is a parallelogram.

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

To prove:

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

Proof:

In ΔDMC and ΔNMB

∠DMC =∠NMB      (Vertically opposite angle)

∠DCM =∠NBM       (Alternate angles)  

By AAA- similarity

$\triangle DMC \sim \triangle NMB$

∴ $\frac{DM}{MN} =\frac{DC}{BN}$

Hence proved.