Let ABCD parallelogram. The inscribed circle in triangle ABD is tangent to BD in E. Show that $$\frac{r_{DEC}}{r_{BEC}}=\tan (\frac{1}{2}\angle ACD)\cdot \tan (\frac{1}{2} \angle ADB)$$ What I have tried is to make some substitutions $$r_{DEC}=r_{1}$$ and $$r_{BEC}=r_{2}$$ so that $$r_{1}=\frac {S_{1}}{p_{1}}$$ and with $$r_{2}$$ the same. But now I'm stuck. I need some help or hint. Thank you.
• The stated relation doesn't appear to be true. Consider the limiting case of $ABCD$ being a square, then LHS is 1 but RHS is $\tan^2{\frac{\pi}{8}}$. – Poon Levi Jan 21 at 10:28
Let $$DC=AB=b$$, $$AD=BC=c$$ and $$BD=a$$ and angles in front these sides in $$\Delta ADB$$ or in $$\Delta CND$$ they are $$\beta$$, $$\gamma$$ and $$\alpha$$ respectively.
Thus, $$\frac{r_{\Delta DEC}}{r_{\Delta ECB}}=\frac{\frac{2S_{\delta DEC}}{DC+DE+EC}}{\frac{2S_{\Delta ECB}}{BC+EB+EC}}=\frac{DE(BC+EB+EC)}{BE(DC+DE+EC)}=$$ $$=\frac{\frac{a+c-b}{2}\left(c+\frac{a+b-c}{2}+EC\right)}{\frac{a+b-c}{2}\left(b+\frac{a+c-b}{2}+EC\right)}=\frac{a+c-b}{a+b-c}=$$ $$=\frac{\sin\alpha+\sin\gamma-\sin\beta}{\sin\alpha+\sin\beta-\sin\gamma}=\frac{2\sin\frac{\alpha+\gamma}{2}\cos\frac{\alpha-\gamma}{2}-2\sin\frac{\beta}{2}\cos\frac{\beta}{2}}{2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}-2\sin\frac{\gamma}{2}\cos\frac{\gamma}{2}}=$$ $$=\frac{\cos\frac{\beta}{2}\left(\cos\frac{\alpha-\gamma}{2}-\cos\frac{\alpha+\gamma}{2}\right)}{\cos\frac{\gamma}{2}\left(\cos\frac{\alpha-\beta}{2}-\cos\frac{\alpha+\beta}{2}\right)}=\frac{\cos\frac{\beta}{2}\sin\frac{\alpha}{2}\sin\frac{\gamma}{2}}{\cos\frac{\gamma}{2}\sin\frac{\alpha}{2}\sin\frac{\beta}{2}}=\frac{\tan\frac{\gamma}{2}}{\tan\frac{\beta}{2}}=$$ $$=\tan\frac{1}{2}\measuredangle ABD\cot\frac{1}{2}\measuredangle ADB.$$