A geometry problem by Yahia El Haw

The area of triangle $ADC$ is given by 0.5 X base X height $= \frac{1}{2} \times 7 \times 2 = 7$ .

The area of parallelogram $ABCD$ is twice that of triangle $ADC$ , hence it is equal to $7 \times 2 = 14$ .

The area of parallelogram $ABCD$ is also equal to base X height $= 6 \times$ "perpendicular distance from $D$ to $AB$ ". Hence, this distance is equal to $\frac{ 14 } { 6} = \frac{7}{3}$ .

Let $x$ be the perpendicular distance from $D$ to $AB$ .

$\text{Area [ADC] = Area[ACB] = } \dfrac{1}{2}(2)(7)=7$

$\text{Area[ABCD] = 7 + 7 = 14}$

So we have,

$14 = 6x$

$x=\dfrac{14}{6}=\boxed{\dfrac{7}{3}}$

$\sigma$ $[ADC]$ = $\frac { 2\cdot 7 }{ 7 }$ $;$ $\sigma$ $[ABCD]$ = $2\cdot 7$ = $14$ = $6$ $\left\| DF \right\|$ $\Longrightarrow$ $\left\| DF \right\|$ = $\frac { 14 }{ 6 }$ = $\frac { 7 }{ 3 }$ .