# Ratio of areas of triangles

There's this geometry problem I tried to solve, but I am not sure if my solution is the easiest one. First I'll explain the problem:

Let's say we have a triangle ABC, a point X in the third of side AB closer to A, a point Y in the third of XB closer to X and then a point Z on BC placed in such a way so that the angles ∠XCB and ∠YZB are the same. What is the ratio of the areas of triangles ABC and XZC.

The way I solved this is by having a triangle ABC in which the height of the triangle passes through point X and its height is equal to XB. Because we know that triangles XCB and YZB share 1 angle and one of the same size (∠ABC and ∠XCB/∠YZB) we can, therefore, say that ∠BYZ=∠BXC=90°. Make a point H on XC, through which the height of XZC passes. That means that the angle ∠XHZ=90°. We can prove that the length of XY equals HZ, by looking at the fact that lines XH and YZ are parallel and because of the fact that the angle ∠XHZ=90° then also lines XY and HZ are parallel which makes XYZH a rectangle. So now we know that XY=HZ. We can now calculate the areas:

$$\text{A of ABC} = {\dfrac{1\cdot\dfrac{2}{3}}{2}} = {\frac{1}{3}}$$ $$\text{A of XZC} = {\dfrac{\dfrac{2}{3}\cdot{\dfrac{2}{9}}}{2}} = {\frac{2}{27}}$$ $$\text{Ratio} = {\frac{1}{3}}\div{\frac{2}{27}} = 4.5$$

Is there an easier way of calculating this?