Let $ABC$ be an acute triangle with $AB<AC$. Let then • $D$ be the foot of the bisector of the angle in $A$, • $E$ be the point on segment $BC$ (different from $B$) such that $AB=AE$, • $F$ be the point on segment $BC$ (different from $B$) such that $BD=DF$, • $G$ be the point on segment $AC$ such that $AB=AG$. Prove that the circumcircle of triangle $EFG$ is tangent to line $AC$.
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Tags: geometry
george_54
07.05.2022 09:40
By LOC it is $BE = 2c\cos B,\operatorname{cosB} = \frac{{{a^2} + {c^2} - {b^2}}}{{2ac}} \Rightarrow BE = \frac{{{a^2} + {c^2} - {b^2}}}{a}$ $CF \cdot CE = (a - 2BD)(a - BE) = \frac{{a(b - c)}}{{b + c}} \cdot \frac{{{b^2} - {c^2}}}{a} = {(b - c)^2} = C{G^2}$ and the result follows.
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Numbertheorydog
07.05.2022 15:15
since $AB=AE$ $\angle ABE=\angle BEA$ similarly,$\angle AEG=\angle AGE=C+\frac{A}{2} $(it is easy to get that $\angle AEG=C+\frac{A}{2})$ We note that $\angle BEA+ \angle AEG+\angle GEF=180^\circ$ which gives $\angle GEF=\frac{A}{2}$ therefore it is enough to prove that $\angle CGF=\frac{A}{2}$ to get $AC$ is tangent to the circle. Which is equivalent to proving that $GF\parallel AD$. since $AD$ is the angle bisector we know, $$\frac{AB}{CA}=\frac{BD}{CD}$$$$AB\cdot CD=CA\cdot BD$$$$-AB\cdot CD=CA\cdot BD-CA\cdot 2BD$$$$-AB\cdot CD=CA\cdot BC-CA\cdot CD-CA\cdot 2BD$$$$CA\cdot CD-AB\cdot CD=CA\cdot BC -CA\cdot 2BD$$$$CD(CA-AB)=CA(BC-2BD)$$$$\frac{CA-AB}{CA}=\frac{BC-2BD}{CD}$$$$\frac{CG}{CA}=\frac{CF}{CD}$$Which establishes that they are parallel.
Tintarn
09.05.2022 12:19
No need for a long computation to prove $FG \parallel AD$: Clearly $DF=DB=DG$ and hence by Thales $FG \perp BG$ but of course also $AD \perp BG$.
SilverBlaze_SY
13.04.2024 18:08
Alternate Solution (Synthetic): Let $\angle ABC=\beta, \angle CAB=\alpha, \angle BCA=\gamma$ Join $GD$. According to the problem, $AD$ is an angle bisector. Given, $AB=AG$. Therefore, by SAS congruency, $\Delta ABD\cong \Delta AGD$ $\implies \angle ABD=\angle AGD=\beta$. By some simple angle chasing, we get $\angle BAC=\beta-\gamma$. Also, it is given that $AB=AE \implies \angle ABE=\angle AEB=\beta$. Therefore, as $\angle AEB=\angle AGD$, we get that quadrilateral $AEDG$ is cyclic. Then, $\angle EAG=\angle GDC$. Due to congruency, $AB=AD=AG$ and $GD=BD=DF$ and $\Delta AEG\cong\Delta DFG$. Therefore, $\boxed{\angle AEB=\angle AGD}, \boxed{\angle AEG=\angle DFG=\angle DGF}$ Then: $$\angle GEC=180^{\circ}-(\angle AEB+\angle AEG)=180^{\circ}-(\angle AGD+\angle DGF)$$$$\boxed{\angle GEC=\angle FGC}$$And by the converse of alternate segment theorem, line $AC$ is tangent to the circumcircle of $\Delta EFG$, and we're done!
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