Incircle Frustation

Geometry Level 3

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

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

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

where rcr^c is the radius of the incircle of ΔABC\Delta ABC.

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

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