In a parallelogram $ \mathrm{ABCD}, \mathrm{E} $ and $ \mathrm{F} $ are the mid-points of sides $ \mathrm{AB} $ and $ \mathrm{CD} $ respectively (see below figure). Show that the line segments AF and $ \mathrm{EC} $ trisect the diagonal BD. "
Given:
In a parallelogram \( \mathrm{ABCD}, \mathrm{E} \) and \( \mathrm{F} \) are the mid-points of sides \( \mathrm{AB} \) and \( \mathrm{CD} \) respectively.
To do: We have to show that the line segments $AF$ and \( \mathrm{EC} \) trisect the diagonal $BD$.
Solution:
$A B C D$ is a parallelogram.
We know that,
Opposite sides of a parallelogram are equal and parallel.
This implies,
$A B \| D C$
$A B=D C$
$\Rightarrow A E \| F C$
$\frac{1}{2} A B=\frac{1}{2} D C$
$\Rightarrow \mathrm{AE} \| \mathrm{FC}$
$\mathrm{AE}=\mathrm{FC}$
Therefore,
$AECF$ is a parallelogram.
This implies,
$\mathrm{AF} \| \mathrm{EC}$
$\Rightarrow \mathrm{EQ} \| \mathrm{AP}$ and $\mathrm{FP} \| \mathrm{CQ}$
In $\triangle B A P$,
$E$ is the mid-point of $A B$ and $E Q \| A P$.
By converse of mid-point theorem,
$Q$ is the mid-point of $B P$.
This implies,
$\mathrm{BQ}=\mathrm{PQ}$...........(i)
In $\triangle D Q C$,
$F$ is the mid-point of $D C$
$F P \| C Q$
By converse of mid-point theorem,
$P$ is the mid-point of $D Q$.
Therefore,
$\mathrm{PQ}=\mathrm{DP}$........(ii)
From (i) and (ii), we get,
$B Q=P Q=P D$
Therefore, $CE$ and $AF$ trisect the diagonal $BD$.
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