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.
