In an acute-angled triangle $ABC$, point $M$ is the midpoint of side $BC$ and the centers of the $M$- excircles of triangles $AMB$ and $AMC$ are $D$ and $E$, respectively. The circumcircle of triangle $ABD$ intersects line $BC$ at points $B$ and $F$. The circumcircle of triangle $ACE$ intersects line $BC$ at points $C$ and $G$. Prove that $BF\hspace{0.25mm} = \hspace{0.25mm} CG$ .
Problem
Source:
Tags: MMO 2019, Macedonia, geometry, circumcircle
21.04.2019 03:06
We'll prove the problem using the following two well known properties: $(1)\hspace{2mm} If\hspace{1mm} I\hspace{1mm} is\hspace{1mm} the\hspace{1mm} incenter\hspace{1mm} of\hspace{1mm} ABC\hspace{1mm}, then\hspace{1mm} \measuredangle BIC\hspace{0.5mm} =\hspace{0.5mm} \frac{\measuredangle BAC}{2} + 90^{\circ}\hspace{1mm}, \\ (2)\hspace{2mm} If\hspace{1mm} I\hspace{1mm} is\hspace{1mm} the\hspace{1mm} incenter\hspace{1mm} of\hspace{1mm} ABC\hspace{1mm}, and\hspace{1mm} I_{A}\hspace{1mm} is\hspace{1mm} the\hspace{1mm} A-excenter\hspace{1mm} of\hspace{1mm} ABC\hspace{1mm},\hspace{1mm} then\hspace{1mm} IBI_{A}C\hspace{1mm} is\hspace{1mm} cyclic\hspace{1mm}.$ Let the circumcircle of $ABD$ intersect $AM$ at $A$ and $K$, let the circumcircle of $ACE$ intersect $AM$ at $A$ and $L$, let $I_{C}$ be the incenter of $ABM$, and let $I_{B}$ be the incenter of $ACM$. Using $(1)\hspace{1mm}$ and $(2)$, we get $\measuredangle BKM\hspace{1mm} =\hspace{1mm} 180^{\circ}\hspace{1mm} -\hspace{1mm} \measuredangle BKA\hspace{1mm} = \hspace{1mm} \measuredangle BEA\hspace{1mm} =\hspace{1mm} 180^{\circ}\hspace{1mm} -\hspace{1mm} \measuredangle BI_{C}A\hspace{1mm} =\hspace{1mm} 90^{\circ}\hspace{1mm} -\hspace{1mm} \frac{\measuredangle BMA}{2}\hspace{1mm} =\hspace{1mm} 90^{\circ}\hspace{1mm} -\hspace{1mm} \frac{\measuredangle BMK}{2}$, so $\measuredangle MBK\hspace{1mm} =\hspace{1mm} 180^{\circ}\hspace{1mm} -\hspace{1mm} \measuredangle BKM\hspace{1mm} -\hspace{1mm} \measuredangle BMK\hspace{1mm} =\hspace{1mm} \measuredangle BKM $, i.e. $MK\hspace{1mm} =\hspace{1mm} MB\hspace{1mm} =\hspace{1mm} MC\hspace{1mm}$, Similarly, we get $ML\hspace{1mm} =\hspace{1mm} MC\hspace{1mm} =\hspace{1mm} MB$, and therefore, $K\hspace{1mm} \equiv \hspace{1mm} L$, i.e. the second intersection of the circumcircles of $ABD$ and $ACE$ lies on $AM$. Therefore, $M$ lies on the radical axis of $(ABD)$ and $(ACE)$, i.e. $MB\hspace{1mm} \cdot \hspace{1mm} MF\hspace{1mm} =\hspace{1mm} MC\hspace{1mm} \cdot \hspace{1mm} MG\hspace{1mm}$, and since $MB\hspace{1mm} =\hspace{1mm} MC$, we get $MF\hspace{1mm} =\hspace{1mm} MG$, i.e. $MB\hspace{1mm} +\hspace{1mm} BF\hspace{1mm} =\hspace{1mm} MC\hspace{1mm} +\hspace{1mm} CG$, and using the fact that $MB\hspace{1mm} =\hspace{1mm} MC$ yet again, we get $BF\hspace{1mm} =\hspace{1mm} CG$, as desired.
21.04.2019 05:29
This problem is closed to here
18.10.2019 14:16
It also can be solved by using bary bash
23.07.2020 18:51
I posted solution here: https://artofproblemsolving.com/community/c6h1833948p12289974