Area Properties Of Triangles

Let's split area $x$ from the vertex, such that $x = a + b$ .

We will use the fact that for triangles with the same height (or base), the ratio of their areas is the ratio of their base (or height).

Triangles $AXD$ and $BXD$ along base $AB$ have identical heights. Therefore, their areas must in the ratio of their bases:

$a:5 = AD : DB$

Triangles $ACD$ and $BCD$ along base $AB$ have identical heights. Therefore, their areas must be in the ratio of their bases:

$a+b+8 : 5+10 = AD:DB$

Since these 2 sets of triangles have the same base, we conclude that

$a: 5 = AD : DB = (a+b+8) : 15 \Rightarrow b = 2a - 8$

We repeat this process on the other side.

Triangles $AXE$ and $CXE$ along base $AC$ have identical heights. Therefore, their areas must be in the ratio of their bases:

$b:8 = AE : EC$

Triangles $ABE$ and $CBE$ along base $AC$ have identical heights. Therefore, their areas must be in the ratio of their bases:

$a+b+5 : 8 + 10 = AE : EC$

Since these 2 sets of triangles have the same base, we conclude that

$b:8 = AE: EC = (a+b+5) : 18 \Rightarrow 5b = 4a + 20$

Solving these 2 equations, we obtain $5 ( 2b-8) = 4a+20 \Rightarrow a = 10, b = 2 \times 10 - 8 = 12$ .

Hence, $x = a + b = 10 + 12 = 22$

A complete discussion of the solution to this problem, including proofs of the theorems used, can be found here .

I don't know if the author of this work invented the names "Crossed Ladder Theorem" and "Extended Crossed Ladder Theorem" but it looks like Menelaus's Theorem is embedded in there !!

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 area of the yellow region is $a+b=10+12=$ $\boxed{22}$

Please refer to Patryk's diagram. Tr. DXE = 4 unit² because Tr.DBX / Tr.CBX = 5/10 = 1/2. Now apply Menelaus's Theorem to Tr. ADC with BXE as the transversal to get: (12/(x-4)) ((x+5)/9) (1/2)=1 which yields x = 22 unit²