ABC is a triangle. The bisector of the exterior angle at B and the bisector of angle C intersect each other at D. Prove that angle D is half of angle A.

Asked by srikar1009 | 1st Sep, 2010, 12:00: AM

Expert Answer:

Let the bisector of angle C meets side AB of triangle ABC at E.
Let the exterior angle at B be 2x, and angle C be 2y,
Angle B = 180 -  2x and Angle A = 180 - (180 - 2x) - 2y = 2x - 2y       i.e. 180 - B - C.
In ΔBEC, Angle BEC = 180 - Angle B - C/2 = 180 - (180 - 2x) - y = 2x - y
In ΔBDE, Angle BED = 180 - Angle BEC = 180 - 2x + y
And Angle D = Angle BDE = 180 - Angle DBE - Angle DEB = 180 - x - (180 - 2x + y) = x - y
As we see Angle D = x - y = Angle A/2 i.e (2x - 2y)/2.
Regards,
Team,
TopperLearning.

Answered by  | 1st Sep, 2010, 10:25: PM

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