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.
Answers
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
