Let $M$ be the midpoint of side BC of an acute-angled triangle $ABC$. Let $D$ and $E$ be the center of the excircle of triangle $AMB$ tangent to side $AB$ and the center of the excircle of triangle $AMC$ tangent to side $AC$, respectively. The circumscribed circle of triangle $ABD$ intersects line$ BC$ for the second time at point $F$, and the circumcircle of triangle $ACE$ is at point $G$. Prove that $| BF | = | CG|$.
Problem
Source: 2020 Estonia TST 3.2
Tags: geometry, excircle, equal segments, circumcircle
jacoporizzo
19.11.2020 03:35
Let $\angle{BAM}=\alpha, \angle{CAM}=\beta$. We proceed to find $\angle{BAF}$. It is known that the excenter is the intersection of exterior angle bisectors, so $$\angle{DAB}=90-\frac{\alpha}{2}, \angle{ABD}=90-\frac{B}{2} \Rightarrow \angle{BDA}=\frac{\alpha + B}{2}$$Thus, $$\angle{BAF}=180-\angle{AFB}-\angle{FBA}=180-(180-\frac{\alpha + B}{2})-B=\frac{\alpha-B}{2}$$Then by LoS on $ABF$, $$\frac{c}{\sin{(\frac{\alpha+B}{2})}}=\frac{BF}{\sin{(\frac{\alpha-B}{2})}} \Rightarrow BF = \frac{c\sin{(\frac{\alpha-B}{2})}}{\sin{(\frac{\alpha+B}{2})}}$$and similarly $CG=\frac{b\sin{(\frac{\beta-C}{2})}}{\sin{(\frac{\beta+C}{2})}}$, so it is enough to show $$b\sin{(\frac{\beta-C}{2})}\sin{(\frac{\alpha+B}{2})}=c\sin{(\frac{\alpha-B}{2})}\sin{(\frac{\beta+C}{2})}$$After simplifying using product-sum, we want $$b(\sin{C}-\sin{\beta})=c(\sin{B}-\sin{\alpha})$$But from LoS on $ABC$ we get $b\sin{C}=c\sin{B}$, and by Ratio Lemma on $ABC$ with cevian $AM$ we have $b\sin{\beta}=c\sin{\alpha}$, and subtracting the two gives us the desired result.
MiltonMath12
19.11.2020 03:38
parmenides51 wrote: Let $M$ be the midpoint of side BC of an acute-angled triangle $ABC$. Let $D$ and $E$ be the center of the excircle of triangle $AMB$ tangent to side $AB$ and the center of the excircle of triangle $AMC$ tangent to side $AC$, respectively. The circumscribed circle of triangle $ABD$ intersects line$ BC$ for the second time at point $F$, and the circumcircle of triangle $ACE$ is at point $G$. Prove that $| BF | = | CG|$. The excircles are the M-excircles?