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`        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`
2 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
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2 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
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2 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.
```
2 years ago
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