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## Triangle – Circle problem

Given triangle ABC. $\angle C = 90^\circ$ . CD is height of triangle ( point D is on hypotenuse AB ). The radiuses of incircles of triangles ACD and ABD are $r_1 \: and \: r_2$ . Find the radius of incircle of triangle ABC .

Solution :

Let $O_1$ be center of incircle of triangle ACD with radius $r_1$ and $O_2$ be center of incircle of triangle BCD with radius $r_2$ . Let the center of incircle of triangle ABC $O_3$ with radius R. ( we have to find R ) And let AC=b, CB=a, AB=c .

The center of incircle is the intersection point of bisectors of triangle.

$\triangle ACO_1 \sim \triangle BCO_2 \sim \triangle ABO_3$

$\frac{ r_1 }{R} = \frac{b}{c} \: and \: \frac{ r_2 }{R} = \frac{a}{c}$

$\frac{ r^2_1 + r^2_2 }{ R^2 } = 1 \Rightarrow R= \sqrt{ r^2_1 + r^2_2 }$

$CD = R + r_1 + r_2$

1. October 1, 2011 at 14:46
2. October 4, 2011 at 14:13
3. October 31, 2011 at 11:43
4. November 6, 2011 at 16:28