circular geometry question (1 Viewer)

[jenslekman]

rate the difficulty of question and please help me do it

BE and CF are altitudes of triangle ABC. M is the midpoint of BC. Prove that angle FME = 180 - 2(angle BAC)

 

[barbernator]

Draw a GOOD diagram so you can follow my reasoning, i'm not sure if it's the most efficient answer, but it works.

i am leaving out the angle symbol, but u get the point.

CEF+BFE=2EAF+AFE+AEF (exterior angle of triangle)
CEF+BFE=2EAF+ECB+FBC (exterior angle of a cyclic quadrilateral)(EFBC is cyclic, as CEB and CFB are angles in a semicircle)
CEM+MEF+BFM+MFE=2EAF+ECB+FBC (Splitting up the angles into parts)
therefore, MEF+MFE=2EAF (angles cancel)
therefore 180-EFM=2EAF (angle sum of triangle EFM)
therefore EFM=180-2EAF as required

Took me a while to get this one, so i think it was quite hard. Probably an easier method to solving

 

[apollo1]

heres another way

since BE and CF are altitudes of triangle ABC. they are perpendicular to their respective sides. therefore since the line BC subtends two equal angles on its same side the points F, E, B and C are concyclic.
and since angle BFC = angle BEC = 90 BC is the diameter of the circle through those points with M centre.
now in triangle BAE, let angle BAC = theta. therefore since angle BEA = 90, angle ABE = 90-theta (angle sum of a triangle)
therefore since angle FME = 2angle FBE (angle at centre twice angle at circumference subtended by same arc), angle FME = 2(90-theta)

hence angle FME = 180 - 2angle BAC.

 

[barbernator]

heres another way

since BE and CF are altitudes of triangle ABC. they are perpendicular to their respective sides. therefore since the line BC subtends two equal angles on its same side the points F, E, B and C are concyclic.
and since angle BFC = angle BEC = 90 BC is the diameter of the circle through those points with M centre.
now in triangle BAE, let angle BAC = theta. therefore since angle BEA = 90, angle ABE = 90-theta (angle sum of a triangle)
therefore since angle FME = 2angle FBE (angle at centre twice angle at circumference subtended by same arc), angle FME = 2(90-theta)

hence angle FME = 180 - 2angle BAC.

nice one! that diagram had a lot going on.

 

[jenslekman]

both of you are geniuses! cant believe i didnt see that @#$%^&* i looked at it for an hour-ish and i picked up the cyclic quadrilaterals but not the angle at centre =S Thanks for your solutions!