Let ABC be a triangle in which angle A = 60 degrees. Let BE and CF be the bisectors of the angles B and C, with E on AC and F on AB. Let M be the reflection of A in the line EF. Prove that M lies on BC.
How do I start?
49General Inferences!!
E is equidistant from AB and AC
F is equidistant from AC and BC
perp from E to AB (EX), perp from F to AC (FY) and perp from A to EF (AZ) are concurrent!
B+C=120°
aur dikha toh likh dunga!! :P
1F is equidistant from AC and BC
Is this some property of angle bisectors?
perp from E to AB (EX), perp from F to AC (FY) and perp from A to EF (AZ) are concurrent!
How did you infer that they are concurrent?
B+C=120°
I did not get how you concluded the answer from this :|
71How did you infer that they are concurrent?
Perpendicular bisectors are concurrent!
1@Vivek - If you understood what subho meant can you explain the rest -
F is equidistant from AC and BC & B+C=120° ( I know how they sum up to 120 but how do you infer from this that the image of A lies in BC