, 2011

Let $F'$ be the point of intersection of the bisectors of the angles $\angle ABD$ and $\angle ADC$. Since $\angle F'BD = \angle ABD/2$ and $\angle F'DC = (\angle ABD + \angle BAD)/2$, we have $\angle I_1F'D = \angle BF'D = \angle BAD/2 = \angle I_1AD$. Therefore $A, I_1, D, F'$ are concyclic. We also have $\angle IAI_2 = (\angle BAC - \angle DAC)/2 = \angle BAD/2 = \angle BF'D = \angle IF'I_2$, and hence $A, I, I_2, F'$ are concyclic too. We conclude that $F' = F$. Similarly, $E$ is the point of intersection of the bisectors of $\angle ACD$ and $\angle ADB$.

Since $\angle IAI_2 = \angle I_1AD$ and $\angle AI_2I = \angle AI_2E = \angle ADE = \angle ADI_1$, the triangles $IAI_2$ and $I_1AD$ are similar. Hence $II_2/I_1D = AI_2/AD$. Similarly, $II_1/I_2D = AI_1/AD$. As $AI_1 = AI_2$ these give $II_2 \cdot I_2D = II_1 \cdot I_1D$.

By Menelaus Theorem for the line $FI_1$ and the triangle $EI_2D$, and for the line $EI_2$ and the triangle $FI_1D$ we have
$$\frac{DF}{FI_2} \cdot \frac{I_2I}{IE} \cdot \frac{EI_1}{I_1D} = 1, \quad \text{and} \quad \frac{DE}{EI_1} \cdot \frac{I_1I}{IF} \cdot \frac{FI_2}{I_2D} = 1,$$
respectively. From these we obtain
$$\frac{EI_1^2}{FI_2^2} = \frac{I_1D \cdot I_1I}{I_2D \cdot I_2I} \cdot \frac{EI \cdot ED}{FI \cdot FD} = \frac{EI \cdot ED}{FI \cdot FD}.$$