Find the missing area

Split the area labelled X into two areas as shown in the above figure, with areas $a$ and $b$ . It follows that

$\dfrac{(8 + a)}{b} = \dfrac{10}{5} = 2$

and

$\dfrac{ (5 + b)}{a} = \dfrac{10}{8} = \dfrac{5}{4}$

From the above two equations, we get

$a - 2 b = -8$

and

$5 a - 4 b = 20$

Solving this system of equations, results in $a = 12$ and $b = 10$ , Hence $x = a + b = 12 + 10 = 22$ .

Consider the diagram. Recall that the areas of triangles with equal altitudes are proportional to the bases of the triangles. We have, $\dfrac{AD}{DB}=\dfrac{A_{CDA}}{A_{CDB}}=\dfrac{A_{XDA}}{A_{XDB}}$

$\dfrac{a+b+8}{5+10}=\dfrac{a}{5}$ $\implies$ $5(a+b+8)=15a$ $\implies$ $a+b+8=3a$ $\implies$ $b+8=2a$ $\color{#D61F06}(1)$

$\dfrac{AE}{EC}=\dfrac{A_{ABE}}{A_{BEC}}=\dfrac{A_{AXE}}{A_{EXC}}$

$\dfrac{a+b+5}{8+10}=\dfrac{b}{8}$ $\implies$ $8(a+b+5)=18b$ $\implies$ $4a+4b+20=9b$ $\implies$ $4a+20=5b$ $\color{#D61F06}(2)$

Substitute $\color{#D61F06}(1)$ in $\color{#D61F06}(2)$ , we have

$2(b+8)+20=5b$ $\implies$ $2b+16+20=5b$ $\implies$ $36=3b$ $\implies$ $12=b$

It follows that,

$2a=b+8=12+8=20$ $\implies$ $a=10$

Finally, the desired area is $a+b=10+12=$ $\boxed{22}$

NOTE: My diagram is no true to scale.

The first stage of this solution could probably be avoided but I am yet to find the shortcut.

Areas: $\triangle ADE=8$ , $\triangle BDF=5$ and $\triangle ABD = 10$ . Let $DE=a$ and $DF=b$ .

Since $\triangle ABD$ and $\triangle ADE$ share the same height (perpendicular to $EB$ ), then $DB$ must be equal to $\cfrac{10}{8}\ a=\cfrac{5}{4}\ a$ .

Using similar logic on $\triangle ABF$ , you can show that $AD=2b$ .

Since $\angle ADB=\angle EDF$ , we can use the ratios between the sides of $\triangle ABD$ and $\triangle DEF$ to work out the area of $\triangle DEF$ . $DE\times DF =ab$ , whilst $AD\times BD=\cfrac{5}{4}a\times 2b= \cfrac{5}{2}ab\implies \triangle DEF \times \cfrac{5}{2}=\triangle ABD = 10 \implies \triangle DEF = 4$ .

Let the heights of triangles $\triangle ADE,\ \triangle ACF,\ \triangle BDF,$ and $\triangle BCE$ be $h_1, h_2, h_3$ and $h_4$ respectively. Then define the ratios $\cfrac{h_1}{h_2}=x$ and $\cfrac{h_3}{h_4}=y$ .

Because they share the same base ( $AD$ ), the area of $\triangle ACD= x\triangle ADE = 8x$ , and so by subtraction the area of $\triangle CDE=8x-8$ . Applying this to $\triangle BDF$ , $\triangle BCD$ and $\triangle CDF$ , you can show also that $\triangle CDF=5y-5$ .

$\triangle CDE$ and $\triangle DEF$ share base $DE$ , so $y\triangle DEF=\triangle CDE=4y$ . $\triangle DEF$ and $\triangle CDF$ share base $DF$ , so $x\triangle DEF=\triangle CDF=4x$ .

Equating the expressions for $\triangle CDE$ gives $4y=8x-8 \implies 5y=10x-10$ and then for $\triangle CDF$ gives $5y-5=4x \implies 5y=4x+5$ .

Thus, $10x-10=4x+5 \implies x=\cfrac{5}{2}$ .

The area of quadrilateral $CEDF = \triangle CDE + \triangle CDF= 8x-8 + 4x =12x-8 = 30-8=\boxed{22}$ .