Why does this proportion work?

[DanaAJames]

In this diagram, triangleXYZ is inscribed into the circles. O is the center of the larger circle. OZ=x, altitude XO=x-5, and OY=x-9. angleXOZ and angleXOY 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 triangleOYX is similar to triangleOXZ? I appreciate any information as to why this works.

 

[pka]

In this diagram, triangleXYZ is inscribed into the circles. O is the center of the larger circle. OZ=x, altitude XO=x-5, and OY=x-9. angleXOZ and angleXOY 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.

In the ratio \(\displaystyle \dfrac{a}{b}=\dfrac{b}{c} \), the number \(\displaystyle b\) is the mean proportional between \(\displaystyle a~\&~c\)

In any right triangle the length of the altitude to the hypotenuse is the mean proportional between the lengths of the two line-segments of the hypotenuse.

Look at your diagram. The length of \(\displaystyle \overline{OX} \), the altitude, is a mean proportional between \(\displaystyle \overline{OY}~\&~\overline{OZ} \).