Question

Please give me statement and Proof of Converse of Midpoint Theorem. Please give appropriate answers. If wrong answers, As usual I will report them so please give appropriate answers.

Answer 1

Answer :

Given : A triangle ABC where E is the midpoint of AB and F is some point on AC and EF is parallel to BC

Statement : The line drawn through the midpoint of one side of a triangle is parallel to the another side bisects the third side .

Construction : Through C draw CM parallel to AB . Extend EF and Mark the point where it cuts CM be D .

Proof :

In quadrilateral EBCD

ED = BC ( given )

EB = CD ( construction )

Since, we have both the pairs of opposite side as parallel and equal it is a parallelogram.

As parallel sides of a parallelogram are equal we have ,

EB = DC

EB = EA ( e is midpoint of AB )

EA = DC - i

Also , EB is parallel to DC ( as AB is parallel to CD by construction ) with transversal ED .

Therefore , Angle AEF = Angle CDF ( alternate angles ) - ii

Now if we observe triangle AEF and triangle CDF ,

Angle AFE = Angle CFD ( v.o.a )

Angle AEF = Angle CDF ( from ii )

AE = CD ( from i )

Therefore ∆ AEF is congruent to ∆ CDF by ASA rule .

Thus , AF = CF by cpct

Hence F is the midpoint of AC

Therefore proved