In a parallelogram $A B C D, E$ and $F$ are the mid-points of sides $A B$ and $C D$ respectively. Show that the line segment $A F$ and $E C$ trisect the diagonal $B D$ .

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Solution

In $△ A P D$ & $△ C Q B$
$∠ P A D = Q C B (E C ∥ A F)$ & $(A D ∥ B C)$
$A D = B C$ (opposite sides of parallelogram)
$∠ C B Q = ∠ A D P$ (corresponding angles for $B C ∥ A D$ )
So, $△ A P D ≅ △ C O B$ (SAS criteria)
So, $△ P = B Q$ (CPCT)
In $△ C Q D$
$D F = \frac{1}{2} (C D)$ (As F is midpoint)
So, $\frac{C D}{D F} = 2$
and also, In $△ D P F$ & $△ D Q C$
$∠ D = ∠ D$ (common)
$∠ D F P = ∠ D C Q$ (corresponding angles as $E C ∥ F A$ )
So, $△ D P F \sim △ D Q C$
So, $\frac{C D}{D F} = \frac{Q D}{P D}$ (BPT)
$2 = \frac{Q D}{P D} = \frac{D P + P Q}{P D} = 1 + \frac{P Q}{P D}$
So, $P Q = P D$
So, $P Q = P D = B Q$
Thus, P,Q trisects BD

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