Online Geometry theorems, problems, solutions, and related topics.
Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.This entry contributed by Ajit Athle.Click the figure below to see the problem 754 details.
http://img100.imageshack.us/img100/2228/problem754.pngLet I is the incenter of triangle DBE ( see sketch)∠DIE= ∠B+∠D/2+∠E/2= 60+ (180-60)/2= 120Since ∠DIE supplement to ∠DOE , quadrilateral DIEO is cyclicO is the intersecting point of angle bisector BI and circumcircle of DIE => O must be excenter of triangle DBELet excircle touch DE at T . We have EQ=ET and DP=DTAnd perimeter of DBE=BP+BQ= AC
Rotate triangle ODB 120(degree) clockwise about O, it becomes triangle OD'B', where D' lies on BC and B'=C. Now since OD=OD', and angle D'OE=120(degree)-angle DOE=60(degree), thus triangle DOE is congruent to triangle D'OE. Hence, DE=D'E. Therefore, perimeter = BE+ED+DB = BE+ED'+D'B' = BC.
Excellent proof Jacob Ha!
Thank you very much!=)
Problem 754Let the point F on the side BC (F is between the points B, C), such that BD=FC.Then Triangle DBO=triangleFCO (OB=OC,DB=FC, <DBO=<FCO=30).So OD=OF and <DOB=<COF.But <COB=120=<DOF.Then <EOF=60=<EOD.So triangleDOE=triangleEOF and DE=EF.SoAC=BC=BE+EF+FC=BE+ED+DB.APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE