Question

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

Solution

## ABCD is a parallelogram$$\because AB||CD$$$$\Rightarrow AE||CF$$ & $$AB=CD$$$$\dfrac{1}{2}AB=\dfrac{1}{2}CD\Rightarrow AE=OF$$in $$AECF$$$$AE||CF$$ adn $$AE=CF$$One pair of opposite sides is equal to $$11$$AECF is a parallelogram$$\because AF||CF$$$$\Rightarrow PF||CQ$$ and $$AP||EQ$$$$\Delta DQC$$                                                                   $$\Delta ABP$$F is the mid point of DC & $$PF||CQ$$                     E is the mid point of AB and $$AP||EQ$$P is the mid point od DQ                                        Q is the mid point of BP$$\Rightarrow PQ=DP$$                                                       $$PQ=QB$$$$PQ=DP=BQ$$Hence the line segment AE & EC triset the diagonal BD.Mathematics

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