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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   
1060326_ed3f6178953046e29c83715ceb3c82e5.png


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.

1205264_1060326_ans_ff8ed2a7ebde42a3a536f3f442e0cb2d.png

Mathematics

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