Semicircles

Online, I saw an interesting problem.

Here’s my solution

By the tangent-secant theorem,

$a^2=b(b+c)$

Set

$L=b+c$

therefore

$a^2=bL$

observe

$b = z \cdot cos(\theta)$

substituting

$a^2=z \cdot L \cdot cos(\theta)$

observe

$z=L \cdot cos(\theta)$

therefore

$L = \frac{z}{cos(\theta)}$

substituting

$a^2=z \cdot \frac{z}{cos(\theta)} \cdot cos(\theta)$

simplifying

$a^2=z^2$

therefore, finally

$\boxed{a=z}$

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I built a GeoGebra app to illustrate the geometry; in GeoGebra, click on Point P (or use your finger) to move it up and down.