Let ABC be an acute triangle and AD an altitude. The angle bisector of ∠DAC intersects DC at E. Let F be a point on AE such that BF is perpendicular to AE. If ∠BAE=45, find ∠BFC.
Triangle AFB is an isosceles right triangle.
Angle ABF = 45
Triangle ADC is a right triangle with the same hypotenuse.
Call the intersection of AD and BF = G
Triangles AFG and BDG are similar
Angle GAF = Angle GBD
Angle GAF = Angle CAF from the given bisector
Angle GBD = Angle CAF
Angle ABC = Angle BAC
Triangle ACB is isosceles with vertex C
Call the midpoint of AB = H
H, F and C are colinear and on the perpendicular bisector of AB.
Angle BHF = 45 degrees
Angle BFC = 135 degrees
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Posted by Jer
on 2024-09-20 13:38:41 |
solution
