# Elementary geometry problem – finding one side of rectangle given two parameters

geometry

This is an elementary problem I have been surprisingly finding hard to solve only by using elementary level math (no trigonometry).

The only given parameters are:

1 is the distance from the intersection (between the diagonal and bisector of $$\angle D$$) to the side $$AB$$

8 is the distance from the intersection (between the diagonal and the bisector) to the side $$CB$$.

I need to find the length of the side $$AB$$. How would you solve it, and what's more important (to me) is how would you explain it to a kid who is just starting to learn geometry?

In the picture below, the red and blue right triangles are similar (because of parallel sides) and the purple triangle $$\triangle DOZ$$ is an isosceles right triangle (because $$DO$$ is an angle bisector, making $$\angle ODZ = 45^\circ$$).
Let $$AX = x$$. Then $$DZ = x$$ as well; therefore $$OZ = x$$, because $$\triangle DOZ$$ is isosceles; therefore $$YC = x$$.
Since the red and blue triangles are similar, their legs are in the same ratio: $$AX : XO = OY : YC$$, or $$x : 1 = 8 : x$$. Rearranging, we get $$x^2 = 8$$, or $$x=\sqrt 8$$.
Finally, $$AB = AX + XB = AX + OY = \sqrt8 + 8$$.