crazyfehmy wrote: Let $D$ be a point different from the vertices on the side $BC$ of a triangle $ABC.$ Let $I, \: I_1$ and $I_2$ be the incenters of the triangles $ABC, \: ABD$ and $ADC,$ respectively. Let $E$ be the second intersection point of the circumcircles of the triangles $AI_1I$ and $ADI_2,$ and $F$ be the second intersection point of the circumcircles of the triangles $AII_2$ and $AI_1D.$ Prove that if $AI_1=AI_2,$ then \[ \frac{EI}{FI} \cdot \frac{ED}{FD}=\frac{{EI_1}^2}{{FI_{\color{red}2}}^2}.\] Typo fixed. Regardless of $\color{blue}AI_1 = AI_2$: $p, p_b, p_c$ are semiperimeters of $\triangle ABC, \triangle ABD, \triangle ACD$. $AI_1, AI_2$ are isogonals WRT $\angle IAD$ $\Longrightarrow$ $F$ is B-excenter of $\triangle ABD$ and $E$ is C-excenter of $\triangle ACD$. Incircles $(I_1), (I_2)$ of $\triangle ABD, \triangle ACD$ touch $BC$ at $X_1, X_2$. Excircles $(F), (E)$ of these triangles touch $BC$ at $Z, Y$. Incircle $(I)$ of $\triangle ABC$ touches $BC$ at $X$. $YX_1 = p_b - BD - (p_c - AC) = p - BC = p_c - CD - (p_b - AB) = ZX_2$. $YX = YC - XC = p_c - (p - AB) = p_b - BD$ $ZX = ZB - XB = p_b - (p - AC) = p_c - CD$ $\triangle ABI_1 \sim \triangle FBD,\ \ \triangle ACI_2 \sim \triangle ECD$ $\Longrightarrow$ $\frac{AI_1}{FD} = \frac{BI_1}{BD}, \ \ \frac{ED}{AI_2} = \frac{CD}{CI_2}$ $\frac{EI}{FI} = \frac{CI_2}{BI_1} \cdot \frac{BX_1}{CX_2} \cdot \frac{YX}{ZX} = \frac{CI_2}{BI_1} \cdot \frac{p_b - AD}{p_c - AD} \cdot \frac{p_b - BD}{p_c - CD}$ $\frac{AI_1 \cdot ED \cdot EI}{AI_2 \cdot FD \cdot FI} = \frac{CD}{BD} \cdot \frac{p_b - AD}{p_c - AD} \cdot \frac{p_b - BD}{p_c - CD} = \frac{BD}{CD} \cdot \frac{p_c}{p_b} \cdot \frac{p_c - AC}{p_b - AB} =$ $= \frac{I_1X_1}{DX_1} \cdot \frac {DX_2}{I_2X_2} = \left(\frac{DI_1}{DX_1} \cdot \frac {DX_2}{DI_2}\right)^2 = \left(\frac{EI_1}{YX_1} \cdot \frac {ZX_2}{FI_2}\right)^2 = \left(\frac{EI_1}{FI_2}\right)^2$ If $\color{blue}AI_1 = AI_2$, the conclusion follows.

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yetti wrote: $AI_1, AI_2$ are isogonals WRT $\angle IAD$ $\Longrightarrow$ $F$ is B-excenter of $\triangle ABD$ and $E$ is C-excenter of $\triangle ACD$. Hello yetti, I am impressed with your solution, but I don't understand this point... Would you mind helping me???

yetti wrote: $ \frac{CD}{BD}\cdot\frac{p_{b}-AD}{p_{c}-AD}\cdot\frac{p_{b}-BD}{p_{c}-CD}=\frac{BD}{CD}\cdot\frac{p_{c}}{p_{b}}\cdot\frac{p_{c}-AC}{p_{b}-AB}$ nice solution! but could someone please help me understand this part? Thanks

happyfun wrote: yetti wrote: $ \frac{CD}{BD}\cdot\frac{p_{b}-AD}{p_{c}-AD}\cdot\frac{p_{b}-BD}{p_{c}-CD}=\frac{BD}{CD}\cdot\frac{p_{c}}{p_{b}}\cdot\frac{p_{c}-AC}{p_{b}-AB}$ nice solution! but could someone please help me understand this part? Thanks Areas of $\triangle ABD, \triangle ACD$ with common A-altitude are in ratio of their bases $BD, CD$. Using Heron formula for these triangle areas, squared, $\frac{p_b(p_b-AB)(p_b-BD)(p_b-AD)}{p_c(p_c-AC)(p_c-CD)(p_c-AD)} = \left(\frac{[ABD]}{[ACD]}\right)^2 = \left(\frac{BD}{CD}\right)^2$ Multiplying the equation by $\frac{CD}{BD} \cdot \frac{p_c}{p_b} \cdot \frac{p_c-AC}{p_b-AB}$ yields $ \frac{CD}{BD} \cdot \frac{p_b-AD}{p_c-AD} \cdot \frac{p_b-BD}{p_c-CD} = \frac{BD}{CD} \cdot \frac{p_c}{p_b} \cdot \frac{p_c-AC}{p_b-AB}$

Bigwood wrote: yetti wrote: $AI_1, AI_2$ are isogonals WRT $\angle IAD$ $\Longrightarrow$ $F$ is B-excenter of $\triangle ABD$ and $E$ is C-excenter of $\triangle ACD$. Hello yetti, I am impressed with your solution, but I don't understand this point... Would you mind helping me??? $\angle IAI_1 = \angle IAB - \angle I_1AB = \frac{_1}{^2}\angle CAB - \frac{_1}{^2}\angle DAB = \frac{_1}{^2}\angle CAD = \angle I_2AD$ $\Longrightarrow$ $AI_1, AI_2$ are isogonals WRT $\angle IAD$. C-excenter $E$ of $\triangle ACD$ is intersection of internal bisector $CI_2I$ of $\angle DCA$ and external bisector $DI_1$ of $\angle ADC$. $I_2E$ is then diameter of circumcircle $\odot(ADI_2)$ $\Longrightarrow$ $E \in \odot(ADI_2)$. $\angle IEI_1 \equiv \angle I_2ED = \angle I_2AD = \angle IAI_1$ $\Longrightarrow$ $E \in \odot (AI_1I)$. Combined, $E$ is also intersection of circumcircles $\odot(ADI_2)$, $\odot(AI_1I)$ other than $A$. Similarly, B-excenter $F$ of $\triangle ABD$ is also intersection of circumcircles $\odot(AI_1D), \odot(AII_2)$ other than $A$.

crazyfehmy wrote: Let $D$ be a point different from the vertices on the side $BC$ of a triangle $ABC.$ Let $I, \: I_1$ and $I_2$ be the incenters of the triangles $ABC, \: ABD$ and $ADC,$ respectively. Let $E$ be the second intersection point of the circumcircles of the triangles $AI_1I$ and $ADI_2,$ and $F$ be the second intersection point of the circumcircles of the triangles $AII_2$ and $AI_1D.$ Prove that if $AI_1=AI_2,$ then \[ \frac{EI}{FI} \cdot \frac{ED}{FD}=\frac{{EI_1}^2}{{FI_2}^2}.\] By some easy angle chasing, we get that $\triangle{ADI_1} \sim \triangle{AI_2I}$ and similarly $\triangle{ADI_2} \sim \triangle{AI_1I}$. By $AI_1=AI_2$, we get that two similarity ratios are equal. So, $\frac{DI_2}{II_1}=\frac{DI_1}{II_2}(1)$. By spiral similarity at $A$ taking $DI_1$ to $I_2I$, we get $E=(AI_1I) \cap (ADI_2)= DI_1 \cap II_2$. Similarly, $F=DI_2 \cap II_1$. By Menelaus' Theorem, we get $\frac{ED}{EI_1}=\frac{FI}{FI_2} \cdot \frac{DI_2}{II_1}(2)$. Similarly, $\frac{FD}{FI_2}=\frac{EI}{EI_1} \cdot \frac{DI_1}{II_2}(3)$. Combining $(1),(2)$ and $(3)$, we get the desired result.

We have $\angle AI_1D = 90 + \frac{\angle BAC}{2}$ and $\angle AII_2 = 90 + \frac{\angle BAC}{2}$. Also $\angle ADI_1 = \angle AI_2I \implies \triangle AII_2 \sim \triangle AI_1D$. Now let $E' = IC\cap DI_1$ and $F'=IB\cap DI_2$. A Spiral similarity centered at $A$ takes segment $II_2$ to $I_1D$ so we get $A=(E'II_1)\cap (E'DI_2) \implies E'=E$. Similarly we get $F'=F$. Now by Menelaus's Theorem we get $\frac{FI_2}{FD} \cdot \frac{DI_1}{I_1E} \cdot \frac{EI}{II_2} = 1$. $\frac{EI_1}{ED} \cdot \frac{DI_2}{I_2F} \cdot \frac{FI}{II_1} = 1$. Dividing these two we get that $\frac{ED}{FD} \cdot \frac{EI}{FI} \cdot \frac{FI_2^2}{EI_1^2} \cdot \frac{DI_1}{II_2} \cdot \frac{II_1}{DI_2} = 1$. By $\triangle AII_2 \sim \triangle AI_1D$ and $\triangle AI_1I \sim \triangle ADI_2$, we have $\frac{DI_1}{II_2} = \frac{AI_1}{AI}$ and $\frac{II_1}{DI_2} = \frac{AI}{AI_2} = \frac{AI}{AI_1}$. So we get that $\frac{ED}{FD} \cdot \frac{EI}{FI} = \frac{EI_1^2}{FI_2^2}$ as desired.