Team Selection Test for IMO 2011

Let $F^{\prime }$ be the point of intersection of the bisectors of the angles $\angle ABD$ and $\angle ADC$. Since $\angle F^{\prime }BD=\angle ABD/2$ and $\angle F^{\prime }DC=(\angle ABD+\angle BAD)/2$, we have $\angle I_1{F}^{\prime }D=\angle B{F}^{\prime }D=\angle BAD/2=\angle I_{1}AD$. Therefore $A,I_1,D,F^{\prime }$ are concyclic. We also have $\angle IA{I}_2=(\angle BAC-\angle DAC)/2=\angle BAD/2=\angle B{F}^{\prime }D=\angle I{F}^{\prime }{I}_2$, and hence $A,I,I_2,F^{\prime }$ are concyclic too. We conclude that $F^{\prime }=F$. Similarly, $E$ is the point of intersection of the bisectors of $\angle ACD$ and $\angle ADB$.



Since $\angle IA{I}_2=\angle I_{1}AD$ and $\angle A{I}_{2}I=\angle A{I}_{2}E=\angle ADE=\angle AD{I}_1$, the triangles $IA{I}_2$ and $I_{1}AD$ are similar. Hence $I{I}_2/I_{1}D=A{I}_2/AD$. Similarly, $I{I}_1/I_{2}D=A{I}_1/AD$. As $A{I}_1=A{I}_2$ these give $I{I}_2\cdot I_{2}D=I{I}_1\cdot I_{1}D$.

By Menelaus Theorem for the line $F{I}_1$ and the triangle $E{I}_{2}D$, and for the line $E{I}_2$ and the triangle $F{I}_{1}D$ we have $$\frac{DF}{F{I}_2}\cdot \frac{I_{2}I}{IE}\cdot \frac{E{I}_1}{I_{1}D}=1,\text{and}\frac{DE}{E{I}_1}\cdot \frac{I_{1}I}{IF}\cdot \frac{F{I}_2}{I_{2}D}=1,$$ respectively. From these we obtain $$\frac{E{I}_1^2}{F{I}_2^2}=\frac{I_{1}D\cdot I_{1}I}{I_{2}D\cdot I_{2}I}\cdot \frac{EI\cdot ED}{FI\cdot FD}=\frac{EI\cdot ED}{FI\cdot FD}.$$