# Incircle Frustation

Geometry Level 2

Let $$\Delta ABC$$ be right-angled, with $$\angle A = 90 ^ \circ$$, & let its circumcircle be $$\tau$$. Let $$\omega$$ be the circle which touches sides $$AB$$ & $$AC$$ & circle $$\tau$$ (internally). Show that radius of $$\omega$$ is given by

$$r = \frac{2bc}{a+b+c}$$,

& find the ratio : $$\frac {r}{r^c}$$

where $$r^c$$ is the radius of the incircle of $$\Delta ABC$$.

Note : You can use the formula for $$r$$ directly, but prove it if you want an extra challenge!

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