$AD$, $BE$, and $CF$ are the bisectors of the interior angles of triangle $ABC$, with $D$, $E$, and $F$ lying on the perimeter. If angle $EDF$ is $90$ degrees, determine all possible values of angle $BAC$.
Problem
Source: USAMO 1987 Problem 2
Tags: geometry unsolved, geometry
Luis González
25.07.2011 09:58
This problem has been discussed many times before, e.g. http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=114998 http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=207066 http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=304972 http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=358388
Arslan
11.12.2022 08:38
Luis González wrote: This problem has been discussed many times before, e.g. http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=114998 http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=207066 http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=304972 http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=358388 Currently, none of the links exists.
Halykov06
17.03.2023 07:11
Very nice problem.İt's my solution. Let $FD$ and $BE$ intersect at point $X$ and let $AD$ and $CF$ intersect at $Y$ .Consider quadirateral $AFDC$.Then we get that $(B,Y;X,E)=-1$.Thus the condition $\angle FDE=90°$ implies $\angle FDA$=$\angle FDB$.Hence F is excenter of the triangle $ADC$.Thus we get $\angle TAF$=$\angle FAD$=$\angle DAC$, where $T$ is intersection of $FD$ and $AC$.Then we get $\angle BAC=120°$