Question

The line segment joining the mid point of two sides of a triangle is parallel to the third side.

Answer 1

REF.Image

Given ABCD is a triangle where E and F are M.P of AB and AC respectively.

We have to prove $EF\parallel BC$.

Construction :- Though C draw a line segment parallel to AB & extends EF to meet line at D.

Proof :- Since $AB\parallel CD$

with transversal ED.

$\angle AEF=\angle CDF$

In $\Delta AEF$ and $\Delta CDF$

$\angle AEF=\angle CDF;\angle AFE=\angle CFD;AF=CF$

$\therefore \Delta AEF\cong \Delta CDF$ (AAS rule)

So EA = DC (CPCT)

But EA = EB $\Rightarrow$ Hence EB = DC.

Now In EBCD , $ED\parallel DC$ & EB = DC

Thus, one pair of opposite sides is equal & parallel

Hence EBCD is a parallelogram

Since opposite sides of parallelogram are parallel

So ; $ED\parallel BC$ i.e, $EF\parallel BC$

Hence proved.