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Today in class me and my friend were discussing cool problems that we've done. And he asked me to.find with proof something interesting. Triangle ABC has right angle at B and we drop a perpindicular from Point B to AC say point D. Then we draw incircles inside the two triangles BDC and BDA. Let the radius of the two incircles be p and q, respectively. How would we go about finding the radius of the incircle ABC.

EDIT:In terms of P and Q

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Let's denote :$\bar {BC} =a , \bar {AB} =c ,\bar {AC} = b$

Note that triangles $BDC ,BDA, ABC $ are similar ,therefore :

$a : p=c : q = b :r = k \Rightarrow a=k\cdot p , c=k \cdot q , b= k\cdot r$

From Pythagorean Theorem we know that :

$b^2=a^2+c^2 \Rightarrow k^2r^2=k^2p^2+k^2q^2 \Rightarrow r^2=p^2+q^2 \Rightarrow r=\sqrt{p^2+q^2}$

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