# In the isosceles triangle, the two squares (white) both have an area of four. Find the area of the shaded.

In the isosceles triangle below, the two squares (white) both have an area of four. Find the area of the shaded. According to my answer key, the answer is $$9\sqrt{2}$$ square units. How can I show the solution through the parallel line theorem?

Let $$x$$ be the distance from the top of the triangle to the middle of the top edge of the top most white square. Let $$y$$ be the distance from the top of the triangle to the centre of the lower white square. By similar triangles we have $$x / 1 = y/ \sqrt{2}$$. Since $$y = x + 2 + \sqrt{2}$$, solving gives $$x = 4 + 3 \sqrt{2}$$. So the height of the whole is isosceles triangle is $$h = 6 + 5\sqrt{2}$$.
Again by similar triangles we can determine that the length of the base is $$b = h(2/x)$$. And so the area of the whole isoceles triangle is $$bh/2 = h^2/x = 8 + 9\sqrt{2}$$.
Since the squares have area $$4$$, they have side $$2$$.
WLOG suppose the bottom-most point on the tilted square has coordinates $$(0,0)$$. Then the coordinates of the right side of the bottom square are $$(\sqrt{2},\sqrt{2})$$ and the coordinates of the upper-rightmost points on the second square are $$(1,2+\sqrt{2})$$. So the line going through them is $$y=\left(-4-3 \sqrt{2}\right) x+5 \sqrt{2}+6$$. Then the height of the triangle is $$6+5\sqrt{2}$$ and its base is $$\frac{2 \left(6+5 \sqrt{2}\right)}{4+3 \sqrt{2}}$$, which imply its area is $$8+9\sqrt{2}$$. Subtracting the area of the squares gives the remaining area as $$9\sqrt{2}$$, as claimed.