Question

Prove that,The segment joining midpoints of any two sides of a triangle is parallel to the third side and half of it.

Answer 1

Step-by-step explanation:

Given:ABCD is a triangle where E and F are mid points of AB and AC respectively.

To Proof: EF||BC

Construction: Throught C draw a line segment parallel to AB and extent EF to meet this line at D.

Proof: Since AB||CD ( By construction )

with transversal ED.

angle AEF = angle CDF ( Alternate angles ).....1

In ∆AEF and ∆CDF

angle AEF = angle CDF ( from (1) )

angle AFE = angle CFD ( vertically opposite angles )

AF=CF ( as F is a mid point of AC )

angle AEF =~ angle CDF ( AAS rule )

So, EA =DC ( CPCT )

But, EA = EB ( E is mid point of AB )

Hence, EB = DC

Now,

In EBCD,

EB||DC and EB=DC

Thus, one pair of opposite side is equal and parallel.

Hence, EBCD is a parallelogram.

Since opposite side of parallelogram are parallel.

So, ED||BC

i.e. EF||BC

Hence,proved

Hope it's helpful....