My solution : Let $AB=c,AC=b,BC=a,AD=u,BD=v$ $u+v=\sqrt{a^2+b^2}$ radius of the inscribe circles on $\triangle {ADC}$ and $\triangle {CDB}$ are equals $\frac{AC+CD}{AD}=\frac{BC+CD}{BD}$ $\frac{b+k}{u}=\frac{a+k}{v}$ $\frac{b^2v+a^2u}{u+v}-uv=k^2$ $b^2\frac{a+b}{a+b+2k}+a^2\frac{b+k}{a+b+2k}=k^2+\frac{(a^2+b^2)(a+k)(b+k)}{(a+b+2k)^2}$ $4k^4+4ak^3+4bk^3+2k^2ab-2a^2b^2-2kab^2-2kba^2=0$ $2(2k^2-ab)(a+k)(b+k)=0$ $\frac{ab}{2}=k^2$

Bourne wrote: radius of the inscribe circles on $\triangle {ADC}$ and $\triangle {CDB}$ are equals $\frac{AC+CD}{AD}=\frac{BC+CD}{BD}$ How did you find that??

Any other solution?

Satyaprakash2009rta wrote: Bourne wrote: radius of the inscribe circles on $\triangle {ADC}$ and $\triangle {CDB}$ are equals $\frac{AC+CD}{AD}=\frac{BC+CD}{BD}$ How did you find that?? this follows as $r=\frac{[ABC]}{s}$ where $s$ is semiperimeter and $r$ is inradius. apply this for both triangles and put them equal to get above relation

Is there any non-computational proof?

Bourne wrote: My solution : Let $AB=c,AC=b,BC=a,AD=u,BD=v$ $u+v=\sqrt{a^2+b^2}$ radius of the inscribe circles on $\triangle {ADC}$ and $\triangle {CDB}$ are equals $\frac{AC+CD}{AD}=\frac{BC+CD}{BD}$ $\frac{b+k}{u}=\frac{a+k}{v}$ $\frac{b^2v+a^2u}{u+v}-uv=k^2$ $b^2\frac{a+b}{a+b+2k}+a^2\frac{b+k}{a+b+2k}=k^2+\frac{(a^2+b^2)(a+k)(b+k)}{(a+b+2k)^2}$ $4k^4+4ak^3+4bk^3+2k^2ab-2a^2b^2-2kab^2-2kba^2=0$ $2(2k^2-ab)(a+k)(b+k)=0$ $\frac{ab}{2}=k^2$ How to get $\frac{AC+CD}{AD}=\frac{BC+CD}{BD}$?