Geometry: Proving length of radius

Question

In the image above, given in circle with center O, BC = 20 and AD = 15, and PQ is parallel to BC, which contains the tangent point F, how can I find the length of the radius?

Answer 1

You use the fact that $APQ$ and $ABC$ are similar and that $PO$, $OQ$, $OF$ and $GD$ are all equal.

You get that $PQ$ is to $BC$ as $AG$ is to $AD$. Also you have that since $PO$, $OQ$, $GD$ and $OF$ are all the same so $PQ$ is twice $GD$. Then you combine this into the equation:

$$2r = b ( 1 - r/h)$$

Where $b$ is the base, $r$ the radius and $h$ the height. Then you solve that for $r$

$$r = {bh \over 2h+b}$$

And insert $b=20$ and $h=15$ and get $r=6$

Answer 2

The area of $\Delta ABC$ is equal to the sum of areas of $\Delta APQ$ and trapezoid $BCQP$: $$A_{\Delta ABC}=\frac12\cdot BC\cdot AD=\frac12\cdot 20\cdot 15;\\ A_{\Delta APQ}=\frac12\cdot PQ\cdot AG=\frac12\cdot 2r\cdot (AD-GD)=\frac12\cdot 2r\cdot (15-r);\\ A_{BCQP}=\frac{BC+PQ}{2}\cdot GD=\frac{20+2r}{2}\cdot r;\\ \frac12\cdot 20\cdot 15=\frac12\cdot 2r\cdot (15-r)+\frac12\cdot (20+2r)\cdot r\Rightarrow \\ 300=50r \Rightarrow \\ r=6.$$