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Suppose we have a right triangle $ABC$ with hypotenuse $AB$. Further, suppose the altitude from $C$ hits hypotenuse $AB$ at point $D$. If the inradius of $ACD$ is $r_1$ and the inradius of $BCD$ is $r_2$, then what is the inradius of triangle $ABC$, in terms of $r_1$ and $r_2$? My guess is that it is $\sqrt{r_1^2+r_2^2}$, but can someone help me confirm or deny this statement?

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  • $\begingroup$ ahh, it appears I have not done enough research... agutie.homestead.com/files/problem/…... but how does one arrive at that? Of course, one obvious method is monstrous algebra bash using the relationships between sids of the right triangle and repeatedly utilizing $A=rs$, but I feel there should be a simple proof for such a simple theorem. $\endgroup$ – Bryce Jan 26 '12 at 22:41
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The triangles $ABC$, $ACD$ and $BCD$ are similar, and so their areas are proportional to the squares of the radii of their in-circles.

Furthermore the area of $ABC$ is equal to the sum of the areas of $ACD$ and $BCD$.

So your guess is correct.

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    $\begingroup$ Minimalist! The OP might like to know that this also proves the Pythagorean Theorem. The areas of the three triangles are $kc^2$, $kb^2$, and $ka^2$ for some $k$, and therefore $kc^2=kb^2+ka^2$. $\endgroup$ – André Nicolas Jan 27 '12 at 0:03

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