Question

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

Answer 1

$AB=CD$ and $AB\parallel CD$(given)

$\frac{1}{2}AD=\frac{1}{2}CD$ and $\frac{1}{2}AB\parallel \frac{1}{2}CD$

$AE=FC$ and $AE\parallel FC$

$\therefore AECF$ is a parallelogram.

In $△DCQ,F$ is the mid-point of $DC$(given)

$FP\parallel CQ$ proved above.

$\therefore P$ is the midpoint of $DQ$(by the converse of mid-point theorem)

i.e., $DP=PQ$ ......$\left(1\right)$

In $△ABP,E$ is the mid-point of $AB$(given)

$EQ\parallel AP$(proved above)

$\therefore Q$ is the midpoint of $BP$ by the converse of mid-point theorem.

$BP=PQ$ ......$\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$

$DP=PQ=BQ$

Hence the line segment $AF$ and $EC$ tri-sect the diagonal $BD$