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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