Show that $ F, A $, and $ I $ are collinear

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Show that $ F, A $, and $ I $ are collinear

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#1 h Dec 15, 2019, 6:09 PM • 1 Y

Y by Adventure10

The inner bisections of the $\angle ABC$ and $\angle ACB$ angles of the $ABC$ triangle are at $I$ . The $BI$ parallel line that runs through the point $A$ finds the $CI$ line at the point $D$ . The $CI$ parallel line for $A$ finds the $BI$ line at the point $E$ . The lines $BD$ and $CE$ are at the point $F$ . Show that $F, A$ , and $I$ are collinear if and only if $AB = AC.$

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#2 h Dec 15, 2019, 9:18 PM • 2 Y

Y by Math5000, Adventure10

Pretty sure this is some IGO problem but I cant remember which.Anway..
By Desargue's Theorm on $ADE$ and $BIC$ we see $DE\parallel BC$ using the given condition.Note that $ADIE$ is a parallelogram and clearly $ADBE$ and $ADEC$ are isosceles trapezoid's.Thus $AE=DB=DI$ and $AD=EC=EI$ .Also $\angle{ADB}=\angle{DAE}=\angle{AEC}$ .Thus $ADB$ and $CEA$ are congruent thus $AB=AC$ . $\square$

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#3 h Dec 17, 2019, 1:56 PM • 3 Y

Y by Math5000, Adventure10, Mango247

First, observe that if $AB=AC,$ we obviously must have $F,A,I$ collinear because then $ADIE$ is a rhombus, meaning that $D,E$ are reflections of each other over $AI.$ Now suppose that $F,A,I$ are collinear. Notice that it suffices to show that $DE||BC$ because then $AI$ must pass through the midpoint of $BC$ by Ceva's Theorem. However, this follows from either homothety/Desargues' on $\triangle ADE$ and $\triangle BIC$ as noted above or a pretty simple angle chase, so we're done. $\blacksquare$

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#4 h Dec 17, 2019, 11:26 PM • 3 Y

Y by Math5000, Adventure10, Mango247

If $AB=AC$ then $F$ is preserved under a reflection about $AI$ , as desired.

If $F, A, I$ are collinear, let $BI \cap AC = M, CI \cap AB = N$ . By Desargues on $\Delta BDN$ and $\Delta CEM$ , we have $BC, MN, DE$ concur. But by Desargues on $\Delta ADE$ and $\Delta IBC$ , we have that $DE \cap BC$ must be the point at infinity along $BC$ . Thus we must have $MN \parallel BC$ , and it is now easy to see that we must have $AB = AC$ (say, by ceva's theorem).

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