AD, the bisector of angle BAC of an acute angle triangle BAC, intersects BC at D . A perpendicular BE is drawn from B on AC at E. D,E are joined.Show that the angle CED is greater than 45 degree.
Let the perpendicular BE intersects AD at O.
An intercept EF equal in length of EO is cut from EA. F,O are joined and prodced to intersect the produced ED at G.
THe triangle EFO is an rt angle isosceles triangle where EO = EF and angle FEO = 90 deg
Hence ang EFG = ang EFO = 45 deg
So the exterior angle of the triangle EFG ie angle CEG = angle EFG + angle FGE.
ie angle CED is greater than 45 deg.
Is there any thing Wrong? if any please comment.