In this diagram, ΔXYZ is inscribed into the circles. O is the center of the larger circle. OZ=x, altitude XO=x-5, and OY=x-9. ∠XOZ and ∠XOY are both right angles. Using the two similar right triangles OYX, and OXZ, this proportion can be written: OY/OX=OX/OZ Then: (x-9)/(x-5)= (x-5)/x
I would like to know why this works. How was this proportion written, why it works and how we know ΔOYX is similar to ΔOXZ? I appreciate any information as to why this works.
Line $YZ$ is a diameter of the inner circle (we're given $\angle XOZ, \angle XOY$ right angles, so $YZ$ is a straight line passing through $O$, the centre of the inner circle). $X$ lies on the circumference of the inner circle. Therefore $\triangle XYZ$ is a right-angle triangle.
If you drop a perpendicular line from the vertex of a right-angle triangle to the hypotenuse, the two new triangles created are right-angle triangles similar to the original triangle, and to each other. When we do this with triangle $\triangle XYZ$, it tells us that $\triangle OYX$ is similar to $\triangle OXZ$. This means the corresponding sides are in the same ratio (e.g. $OY$ in $\triangle OYX$ matches $OX$ in $\triangle OXZ$, $OX$ in $\triangle OYX$ matches $OZ$ in $\triangle OXZ$), i.e. $\dfrac{OY}{OX}=\dfrac{OX}{OZ}$, and so on.