Semicircles

[latexpage]

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.