Proving ∠D=90°-(∠A/2) in Triangle ABC with Bisectors

[Govind_Balaji]

Homework Statement

ABC is a triangle. The bisector of the exterior angle at B and the bisector of ∠C intersect each other at D. Then prove that ∠D=90°-(∠A/2)

Homework Equations

The Attempt at a Solution

I guess I want to prove an untrue thing because by drawing a diagram and working out I get ∠D=∠A/2. I think I have drawn it wrong by mis-understanding the question. Or the question is an error? Or did I work out wrong?

 

[Mentallic]

I get the same result you did and I also oriented the corners of the triangle such that ABC is labelled clockwise instead (C at the top, A at the left), so the intersection point is underneath the triangle and the answer is still the same.

 

[Simon Bridge]

You can do a quick reality check by drawing an arbitrary triangle and constructing the bisectors and measuring the angles. Try it for several triangles.

 

[Govind_Balaji]

You can do a quick reality check by drawing an arbitrary triangle and constructing the bisectors and measuring the angles. Try it for several triangles.

It seems a difficult work. Are there any thing to do with Theoretical geometry.

 

[Govind_Balaji]

Even I tried what Simon said with a simulator called GEOGEBRA. I get the same result.

 

[olivermsun]

I get the same result as you using a + b + c = 180° = (180 - b)/2 + b + a + c/2.

 

[Govind_Balaji]

I guess there must be a printing error in the question like may be the angles changed or like that

 

[Simon Bridge]

That would be my bet.

Aside: do you not know how to bisect and angle using a straight-edge and a compass?
http://www.mathopenref.com/constbisectangle.html

 

[Govind_Balaji]

That would be my bet.

Aside: do you not know how to bisect and angle using a straight-edge and a compass?
http://www.mathopenref.com/constbisectangle.html

Yes I know just lazy to do that because when I miss 1° accuracy, I may get very strange results.