# In a parallelogram ABCD,E and F are the mid-points of sides AB and CD respectively.Show that the line segment AF and EC trisect the diagonal BD

prathmesh lohar
12 Points
5 years ago
GIVEN :- ABCD is a parallelogram where E & F are the mid point of side`s AB&CD respectively to prove :- AF&EC trident BD Proof:- in ABCD AB parallel to CD AE parallel to CD AB=CD AE=CF PROVED
Kochouseph
45 Points
5 years ago
Given: parallelogram ABCD,E and F are the midpoint of AB andCDTo Prove:AF and EC trisect the line BADProo: let the the point of intersection of AF and EC to the diagonal DB be G and H respectively In parallelogram AB =DC 2AE= 2CF AE=CFAlso AE parallel to CF Therefore AECF is a parallelogram Therefore AF parallel to EC In triangle ABGEH parallel to GGAAE =EBTherefore BH =GH. (Midpoint theorem)case 1 In triangle CDH FG parallel to CHDF =CF Therefore DG=GH (midpoint theorem). Case 2From case 1 and case 2DG=GH=BH Hence a proof
C.Vishaal
39 Points
5 years ago
Altight , so let the line segments EC , AF and Dagonal Bd meet at P and Q respectively.

Now, since AB II CD ( Opposite sides of a IIgm), therefore, parts of parallel sides are parallel too, i.e, AE II FC.

Therefore, AFCE is a Parallelogram. ( If in a quadrilateral, a pair of opposite are parallel, it is a parallelogram.) …........{1}

Alright, now take triangle APB.
Since E is the idpt. and EQ II AF, By Coverse of Midpoint Theorem,  too is the midpt. ….......{2}

Similarly, in triangle DQC , F is the midpt and FP II EC, By Coverse of Midpoint Theorem, P too is the midpoint. …........{3}

From {2} and {3}, we see that DP = PQ = QB or in other words, the lines segments AF and EC trisect Diagonal BD.