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

The Proof for

$\dfrac{DM}{MN} =\dfrac{DC}{BN}$    is below

Step-by-step explanation:

Given:

ABCD is a Parallelogram

DC || AB i.e DC || AN

To Prove:

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

Proof:

In  Δ CMD and Δ BMN

∠CDM ≅ ∠MNB     …………..{Alternate angles are equal since DC || AN}

∠CMD ≅ ∠BMN     ……….....{Vertical Opposite Angles are equal}

ΔCMD ≅ ΔBMN     …...........{ By Angle-Angle Similarity test}

If two triangles are similar then their sides are in proportion.

$\dfrac{CM}{BM} =\dfrac{DM}{MN} =\dfrac{DC}{BN}\ \textrm{corresponding sides of similar triangles are in proportion}$  

i.e

$\dfrac{DM}{MN} =\dfrac{DC}{BN}$..........Proved

Answer 2

Answer:

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