Geometry: Find the radius of a circle given parallel secants Suppose a circle has two parallel chords of lengths $a$ and  $b$, and the chords are separated by a distance of $c$. Using only the usual high school geometry theorems (i.e. no trig or calculus), can we derive a formula for the radius? 
I've tried drawing radii in several places without making useful progress. I can't see how to find the radius intersecting the circle and a chord. I can draw the segment from one chord-circle intersection to another, but it need not pass through the center so I can't leverage this to get the radius. 
 A: The perpendicular from the center on these chords will be a perpendicular bisector of the chord. 
Below is an image with a roadmap.

Hope it helps!
A: I just answered this question in chat, so I thought it would be worth posting the answer here.


The horizontal position of a vertical chord of length $c$ in a circle of radius $r$ would be $\pm\sqrt{r^2-\frac{c^2}4}$.
Thus, the distance, $d$, between chords of length $c_1$ and $c_2$ (with $c_1\ge c_2$) would be
$$
d=\sqrt{r^2-\tfrac{c_2^2}4}\pm\sqrt{r^2-\tfrac{c_1^2}4}\tag1
$$
The "$\pm$" depends on whether or not the chords appear on the same side of the center of the circle. In either case, we have
$$
\frac{c_1^2-c_2^2}{4d}=\sqrt{r^2-\tfrac{c_2^2}4}\mp\sqrt{r^2-\tfrac{c_1^2}4}\tag2
$$
To verify $(2)$, simply multiply $(1)$ and $(2)$.
Average $(1)$ and $(2)$ to get
$$
\frac{c_1^2-c_2^2+4d^2}{8d}=\sqrt{r^2-\tfrac{c_2^2}4}\tag3
$$
Thus,
$$
\begin{align}
r^2
&=\frac{\left(c_1^2-c_2^2+4d^2\right)^2+16c_2^2d^2}{64d^2}\tag{4a}\\
&=\frac{c_1^4+c_2^4+16d^4-2c_1^2c_2^2+8c_1^2d^2+8c_2^2d^2}{64d^2}\tag{4b}\\
&=\frac{\left(c_1^2+c_2^2\right)^2+8\left(c_2^2+c_1^2\right)d^2+16d^4-4c_1^2c_2^2}{64d^2}\tag{4c}\\
&=\frac{\left(c_1^2+c_2^2+4d^2\right)^2-4c_1^2c_2^2}{64d^2}\tag{4d}\\
&=\frac{\left((c_1+c_2)^2+4d^2\right)\left((c_1-c_2)^2+4d^2\right)}{64d^2}\tag{4e}
\end{align}
$$
Explanation:
$\text{(4a):}$ solve $(3)$ for $r^2$
$\text{(4b):}$ expand
$\text{(4c):}$ collect
$\text{(4d):}$ collect
$\text{(4e):}$ apply $a^2-b^2=(a+b)(a-b)$
Therefore,
$$
r=\frac{\sqrt{\left((c_1+c_2)^2+4d^2\right)\left((c_1-c_2)^2+4d^2\right)}}{8d}\tag5
$$
In $(5)$, it doesn't matter whether $c_1\ge c_2$ or not.
A: Draw a diagram 

and you may get $\left| \sqrt{r^2-\left(\frac a2\right)^2} \pm \sqrt{r^2-\left(\frac b2\right)^2} \right|=c$
There is plenty of scope for spurious solutions to be introduced here if you try to solve for $r$
If you actually want to find the centre of the circle by geometric construction, take the perpendicular bisectors of the sides of the  isosceles trapezium (trapezoid if you are American).  This should demonstrate that there is one non-spurious solution  
A: $$(c+d)^2 + (a/2)^2 = r^2 = d^2 + (b/2)^2$$
Solve for $d$.

