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Triangle and Incircle

In a setup of right triangles ABC, BDA, and BDC not unlike this


(click on the link, and ignore the written side measures and subtext in that diagram- we don't know any side-lengths here)

The incircle(inscribed circle) of triangle BDA has radius=5 and the incircle of BDC has radius=3. What is the radius of the incircle of ABC?

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marked as duplicate by pedja, LVK, Jennifer Dylan, sdcvvc, J. M. Aug 29 '12 at 10:37

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

The two smaller triangles are similar to each other, and to the big triangle. This is fairly easy to see, the angles match.

What this means informally is that if we put $\triangle ABC$ in a Xerox machine, then with the suitable Reduce setting we get one of the other triangles, and with another Reduce setting, you get the other one.

If $\triangle BDA$ has incircle of radius $p$, and $\triangle BDC$ has incircle of radius $q$, then because one is a scaled version of the other, the hypotenuses of the two triangles are also in the ratio $p:q$. Say the hypotenuses are respectively $kp$ and $kq$.

By the Pythagorean Theorem, the hypotenuse of $\triangle ABC$ is $\sqrt{(kp)^2+(kq)^2}=k\sqrt{p^2+q^2}$.

So the hypotenuses of the three triangles are in the ratio $$p: q: \sqrt{p^2+q^2}.$$

The incircle radii are in the same ratio. So if the two little ones are $p$ and $q$, the big one is $\sqrt{p^2+q^2}$.

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