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

Answer 1

ABCD is a parallelogram

$\because AB||CD$

$\Rightarrow AE||CF$ & $AB=CD$

$\frac{1}{2}AB=\frac{1}{2}CD\Rightarrow AE=OF$

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.