Question

In a parallelogram $ABCD,E$ and $F$ are the midpoints of the sides $AB$ and $DC$ respectively. Show that the line segments $AF$ and $EC$ trisect the diagonal $BD$.

Answer 1

$ABCD$ is a parallelogram where $E$ and $F$ are the mid points of side $AB$ and $CD$ respectively.

$AB||CD$ [opposite sides of a parallelogram are parallel]

$\Rightarrow AE||CF$ [parts of parallel lines are parallel]

and $AB=CD$ [opposite sides of a parallelogram are equal]

$\frac{1}{2}AB=\frac{1}{2}CD$

$\therefore AE=CF$ [given $F$ is a mid-point of $CD$ and $E$ is mid-point of $AB$]

In $△AECF$,

$AE||CF$ and $AE=CF$

One pair of opposite sides is equal and parallel.

$\therefore AECF$ is a parallelogram

$\Rightarrow AF||CE$ [opposite sides of a parallelogram are parallel]

$\therefore PF||CQ$ and $AP||EQ$ [parts of parallel lines are parallel]

In $△DQC$ and $△ABP$,

$F$ is the mid point of $DC$ and $PF||CQ$.

$E$ is the mid point of $AB$ and $AP||EQ$.

Line drawn through mid points of one side of a triangle is parallel to another side, bisects the third side]

$\therefore P$ is the mid point of $DQ$

$\Rightarrow PQ=DP$....(i)

$\therefore Q$ is the mid point of $BP$

$\Rightarrow PQ=QB$.....(ii)

From (i) and (ii),

$DP=PQ=BQ$

Hence, the line segments $AF$ and $EC$ trisect the diagonal $BD$.