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.

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- November 1st 2008, 10:59 PMips_mathsimple plane geometry
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.

- November 8th 2008, 07:10 AMdk_ch
**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.