Prove of inradius of a right angle triangle. R²= R1²+ R2²

Question

A right-angle triangle ABC is present whose right angle is BAC. One perpendicular from point A is taken on BC which is AD. Then in triangle ABC, triangle ABD, and triangle ADC, three inscribed circles are drawn with the radius of r, r1 and r2 respectively. Then prove that r^2 = r1^2 + r2^2Picture Link is attached

Answer 1

In the attached figure we have equalizing areas $$r_1=\frac{CD\times AD}{CD+AD+b}=\frac{bc\cos(\gamma)}{a\cos(\gamma)+a+c}\\r_2=\frac{DB\times AD}{DB+AD+c}=\frac{bc\sin(\gamma)}{a\sin(\gamma)+a+b}\\r=\frac{bc}{a+b+c}$$ so we have to verify that $$\left(\frac{bc\cos(\gamma)}{a\cos(\gamma)+a+c}\right)^2+\left(\frac{bc\sin(\gamma)}{a\sin(\gamma)+a+b}\right)^2=\left(\frac{bc}{a+b+c}\right)^2$$ Since $a\cos(\gamma)=b$ and $a\sin(\gamma)=c$ we have finally to verify that $$(bc\sin(\gamma))^2+(bc\cos(\gamma))^2=(bc)^2$$ We are done.