jred wrote: In⊿ABC,∠A=60°. ⊙I is the incircle of ⊿ABC. ⊙I is tangent to side AB, AC at D, E respectively. Line DE intersects line BI and CI at F, G respectively. Prove that $FG=\frac{BC}{2}$. Using this lemma: $\Delta ABC$, I is the incircle center of $\Delta ABC$, ⊙I is tangent to side AB, AC at D, E respectively. Line DE intersects line BI at F. We can prove $BF \perp CF$. Using this lemma into this problem, we can easily prove $BF \perp CF, BG \perp CG$. Hence, $DGIB,BGFC$ are cyclic quadrilaterals. We have: $\frac{GF}{BC}=\frac{GI}{IB}=\sin \widehat{GBI}=\sin \widehat{IDE}=\sin 30^{\circ}=\frac{1}{2}$ ($\Delta GIF \sim \Delta BIC$) Accordingly, we have $GF=\frac{BC}{2}$ and that's what we need to prove.

Obviously, $\triangle ADE$ is equilateral, also $\widehat{CIF}=\frac{\hat B+\hat C}{2}=60^\circ=\angle DEA$, hence $F,I,C,E$ are concyclic and $CF\bot BF$, similarly $BG\bot CI$. Let $B'=AC\cap BG, C'=AB\cap CF$; due to symmetry, $\angle B'IC=\angle BIC=\angle C'IB=120^\circ$, so $B',C',I$ are collinear. Now let's take $M\in AC| GM\parallel BC$ and $N\in AB|FN\parallel BC$. $GM$ is median in right-angled $\triangle B'GC$, hence $GM=\frac{B'C}{2}=\frac{BC}{2}$. Similarly $FN=GM$; with $GM$ midline in $\triangle BB'C'$, we easily get $\triangle GFN$ isosceles, and $FG=FN$, done. Best regards, sunken rock

How did the obviously equilateral come in?

biomathematics wrote: How did the obviously equilateral come in? Because this triangle is isosceles and one of it'a angle is $60^{\circ}$.

Burii wrote: biomathematics wrote: How did the obviously equilateral come in? Because this triangle is isosceles and one of it'a angle is $60^{\circ}$. Oh oops. I thought D and E were given to be the intersection of the angle bisectors with the corresponding sides

This problem was pretty fun to do, since I haven't been doing geometry recently.

we first prove a lemma. lemma = let $ABC$ be a triangle with incircle touching $AB,AC$ at $D,E$. let $I$ be incentre and let $CI$ intersect $DE$ at $G$. than $\angle CGB = 90$. proof = since $AD=AE$ we get $\angle AED = 90-A/2$ and hence $\angle CEG = 90+A/2$ which gives $\angle CGE =\angle IGE= B/2$ thus $BIGD$ is cyclic quad. and hence , $\angle CGB=\angle IGB = \angle IDG = 90$ thus proving the lemma. main proof = by the above lemma we have $\angle CGB=\angle BFC=90$ and hence $BGFC$ is cyclic quad. with $BC$ as diameter. now let $O$ be midpoint of $BC$ than $\angle FGB=180-\angle FCB = 90+B/2$ thus $\angle FCB = 90-B/2$ similarly $\angle CBG = 90-C/2$ now $\angle OFG = 180-\angle GBC-\angle OFC = 90+C/2-\angle OFC=90+C/2-(180-2\angle FCB)=90+C/2-180-2(90-B/2) = 90-A/2$ thus by sine law in triangle $OFG$ $OB/2sin(90-A/2) =OB/2cosA/2= FG/sin A$ and hence $FG = 2OBsinA/2 = BCsin30 = BC/2$ and hence proved