Partitioning triangles

It is worth noting that this argument can be generalised to give an elegant result.

Using Marvin's ideas, we have $\frac{a}{x} \; = \; \frac{a+b+z}{x+y} \hspace{2cm} \frac{b}{z} \; = \; \frac{a+b+x}{y+z}$ and hence $ay - bx \; = \; xz \hspace{2cm} -az + by \; = \; xz$ and hence $a \; = \; \frac{xz(x+y)}{y^2 - xz} \hspace{2cm} b \; = \; \frac{xz(y+z)}{y^2 - xz}$ If $\Delta$ is the area of the whole triangle, then $\Delta \; = \; a + b + x + y + z \; = \; \frac{y(x+y)(y+z)}{y^2 - xz}$ and hence $\frac{1}{\Delta} + \frac{1}{y} \; = \; \frac{y^2 - xz}{y(x+y)(y+z)} + \frac{1}{y} \; = \; \frac{x+2y+z}{(x+y)(y+z)}$ so that $\frac{1}{\Delta} + \frac{1}{y} \; = \; \frac{1}{x+y} + \frac{1}{y+z}$ a result sometimes called the Ladder Theorem. In this case, therefore, $\frac{1}{\Delta} + \frac{1}{84} \; = \; \frac{1}{120} + \frac{1}{140}$ so that $\Delta = 280$ , which makes the shaded area $280 - 36 - 84 - 56 = \boxed{104}$ .

Let $X$ be the point of intersection of lines inside the triangle. Divide the area of the shaded region into $2$ parts. Name them as $a$ and $b$ . Observed that triangle $AXD$ and triangle $BXD$ share the same altitude from $X$ , so the ratio of their areas is the same as the ratio of their bases. Similarly, triangle $ACD$ and triangle $BCD$ share the same altitude from $C$ , so the ratio of their areas is the same as the ratio of their bases. Moreover, the bases are the same, and thus in the same ratio. As a result, we have:

$\dfrac{a}{36}=\dfrac{a+b+56}{36+84}$

After simplfying the above equation we get

$a=\dfrac{3}{7}b+24$ $\color{#D61F06}(1)$

Applying the same principle to the triangles with the bases $EA$ and $EC$ , we have

$\dfrac{b}{56}=\dfrac{a+b+36}{56+84}$

After simplifying the equation above, we get

$84b=56a+201.6$ $\color{#D61F06}(2)$

Substituting $\color{#D61F06}(1)$ to $\color{#D61F06}(2)$ , we get

$b=56$

It follows that

$a=48$

Finally, the area of the shaded region is $56+48=$ $\color{#D61F06}\boxed{104}$