In a right triangle ABC, $∠ B$ is a right angle and [BD] is an altitude drawn to the hypotenuse [AC]. Prove that |BD| is equal to the sum of the radii of the circles inscribed in the triangles ABC, ADB, and CDB.

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Solution

We know that, in a right angled triangle, radius of incircle $r_{A B C} = \frac{A B + B C - A C}{2}$

Also, ABD is a right triangle with AB as hypotenuse. Its inradius is given by $r_{A B D} = \frac{B D + A D - A B}{2}$

Also, CDB is a right triangle with CB as hypotenuse. Its inradius is given by $r_{C D B} = \frac{C D + B D - C B}{2}$

$r_{A B C} + r_{A D B} + r_{C D B} = \frac{A B + B C - A C + B D + A D - A B + C D + B D - C B}{2}$
$= \frac{2 B D + A D + C D - A C}{2} = B D$ since $A C = A D + D C$

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