.Let $CH$ be an altitude of right-angled triangle $ABC$ ($\angle C = 90^o$), $HA_1$, $HB_1$ be the bisectors of angles $CHB$, $AHC$ respectively, and $E, F$ be the midpoints of $HB_1$ and $HA_1$ respectively. Prove that the lines $AE$ and $BF$ meet on the bisector of angle $ACB$.
Problem
Source: 2021 Sharygin Finals grades X-XI p1
Tags: geometry, concurrency, concurrent, right triangle
SerdarBozdag
01.08.2021 14:05
Cevian nest.
JustinLee2017
13.06.2022 21:01
[asy][asy] import graph; size(6cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -2.96, xmax = 20, ymin = -5.1, ymax = 10.18; /* image dimensions */ pen qqttzz = rgb(0,0.2,0.6); pen xdxdff = rgb(0.49019607843137253,0.49019607843137253,1); /* draw figures */ draw((0,7)--(0,0), linewidth(1)); draw((0,0)--(5,0), linewidth(1)); draw((5,0)--(0,7), linewidth(1)); draw((0,0)--(3.310810810810811,2.364864864864865), linewidth(1)); draw((0,2.9166666666666665)--(3.310810810810811,2.364864864864865), linewidth(1) + qqttzz); draw((3.310810810810811,2.364864864864865)--(2.9166666666666665,0), linewidth(1) + qqttzz); draw((0,7)--(1.6554054054054055,2.6407657657657655), linewidth(1) + xdxdff); draw((3.1137387387387387,1.1824324324324325)--(5,0), linewidth(1) + xdxdff); draw((3.1137387387387387,1.1824324324324325)--(1.926605504587156,1.9266055045871557), linewidth(1) + linetype("2 2") + xdxdff); draw((1.926605504587156,1.9266055045871557)--(1.6554054054054055,2.6407657657657655), linewidth(1) + linetype("2 2") + xdxdff); /* dots and labels */ dot((0,0),linewidth(4pt) + dotstyle); label("$C$", (-0.6,-0.34), NE * labelscalefactor); dot((5,0),dotstyle); label("$A$", (5.08,0.2), NE * labelscalefactor); dot((0,7),dotstyle); label("$B$", (0.08,7.2), NE * labelscalefactor); dot((3.310810810810811,2.364864864864865),linewidth(4pt) + dotstyle); label("$H$", (3.4,2.52), NE * labelscalefactor); dot((0,2.9166666666666665),linewidth(4pt) + dotstyle); label("$A_1$", (-0.66,3.06), NE * labelscalefactor); dot((2.9166666666666665,0),linewidth(4pt) + dotstyle); label("$B_1$", (3,-0.5), NE * labelscalefactor); dot((1.6554054054054055,2.6407657657657655),linewidth(4pt) + dotstyle); label("$F$", (1.74,2.8), NE * labelscalefactor); dot((3.1137387387387387,1.1824324324324325),linewidth(4pt) + dotstyle); label("$E$", (3.2,1.34), NE * labelscalefactor); dot((1.926605504587156,1.9266055045871557),linewidth(4pt) + dotstyle); label("$P$", (2,2.08), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Let $P = AE \cap BF$. Since $AE$ is a median in $\triangle AHB_1$, we have $\frac{\sin \angle HAE}{\sin \angle EAB_1} = \frac{AB_1}{AH} = \frac{AC}{AH+CH}$ Similarly in $\triangle HBA_1$, we have $\frac{\sin \angle HBF}{\sin \angle FBA_1} = \frac{BC}{BH+CH}$. Since $\triangle AHC \sim \triangle CHB$, $\frac{AC}{AH+CH} = \frac{BC}{BH+CH}$. Ceva's theorem finishes. $\quad \square$
R8kt
07.04.2023 13:14
Let N be the midpoint of the angle bisector from C. Then the point of where the lines intersect is the isognal conjugate of N.