Question

In the adjoining figure, ABCD is a trapezium in which AB || DC and M is the mid-point of AD. If MN || DC meets BC at N, prove that N is the mid-point of BC.

Answer 1

Given: ABCD is a trapezium with AB || DC.

M is the midpoint of AD.

MN || DC and MN meets BC at N.

To Prove: N is the midpoint of BC i.e., BN = NC.

Proof (Step-wise):

1. Since MN || DC and AB || DC, we have MN || AB as well all three are parallel.

2. Consider triangle ADC. 

M is the midpoint of AD and the line through M, namely MN, is parallel to DC. 

By the midpoint theorem a line through the midpoint of one side of a triangle parallel to a second side bisects the third side, MN meets AC at its midpoint.

Call that intersection P.

So, AP = PC.

3. Now consider triangle ABC.

P is the midpoint of AC (From step 2) and the line PN (Part of MN) is parallel to AB (From step 1). 

Again, by the midpoint theorem (Applied in this triangle), the line through the midpoint of AC parallel to AB bisects BC.

Hence, PN meets BC at its midpoint.

So, BN = NC.

Therefore, N is the midpoint of BC (BN = NC).