Parallelogram and a Triangle...

We note that the four blue right triangles and the green triangle are similar to $\triangle ABC$ . If the hypotenuse $BC=1$ , then the hypotenuse of a blue triangle is $\frac 14$ and the hypotenuse of the green triangle is $\frac 12$ . Since the area of similar shape figure is directly proportional to the square of the linear dimension. If the area of $\triangle ABC$ is $A_{ABC}=a$ , then the area of a blue triangle is $A_\blue{\text{blue}}\left(\frac 14\right)^2a = \frac 1{16}a$ and the area of the green triangle is $A_\green{\text{green}} \left(\frac 12\right)^2a = \frac 14a$ . Therefore the area of the orange region is:

$\begin{aligned} A_\orange{\text{orange}} & = A_{ABC} - 4 A_\blue{\text{blue}} - A_\green{\text{green}} \\ & = a - \frac 4{16}a - \frac 14a \\ & = \frac 12 a \\ \implies \frac {A_\orange{\text{orange}}}{A_{ABC}} & = \frac {\frac 12 a}a = 0.5 = \boxed{50} \% \end{aligned}$

Since the shapes here are similar the total area can be written as

2A + 2B + 2C = 100%

then to find the shaded area, you can divide by 2 to get

A + B + C = 50%