# Geometry problem, does anyone know how to solve it?

In triangle $\triangle ABC$, angle $\angle B$ is a right angle. $\overline{BD}$ is the altitude, $\overline{DE}$ is perpendicular to $\overline{AB}$ and $\overline{DF}$ is perpendicular to $\overline{BC}$, $a$, $b$, and $x$ are the inradii of the triangles $\triangle AED$, $\triangle DFC$, and $\triangle EDF$ respectively. Prove that $x = \sqrt{ab}$

I don't see any similar triangles I can use here. Do you?

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• $\triangle AED \sim \triangle ABC \sim \triangle DFC$.
• $\triangle ABC \sim \triangle ADB \sim \triangle BDE \sim \triangle EDF$.
• $|AD|:|BD| = |BD|:|DC| \implies |BD|^2 = |AD|\cdot|DC|$.