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

  • Thread starter Thread starter Govind_Balaji
  • Start date Start date
  • Tags Tags
    Angles Geometry
Click For Summary

Homework Help Overview

The problem involves triangle ABC, where the bisector of the exterior angle at B and the bisector of angle C intersect at point D. The goal is to prove that angle D equals 90° minus half of angle A.

Discussion Character

  • Exploratory, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants share their attempts to prove the relationship, noting discrepancies in their results, such as finding angle D equal to angle A/2 instead. Some question whether there is an error in the problem statement or their understanding of the geometry involved.

Discussion Status

Multiple interpretations of the problem are being explored, with participants suggesting that there may be a printing error in the question. Some have attempted practical checks using diagrams and simulations, leading to similar results that challenge the original statement.

Contextual Notes

Participants mention potential issues with the accuracy of their constructions and the complexity of the theoretical geometry involved in the problem.

Govind_Balaji
Messages
83
Reaction score
0

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?
 

Attachments

  • Untitled.png
    Untitled.png
    7.4 KB · Views: 516
Physics news on Phys.org
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.
 
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.
 
Simon Bridge said:
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.
 
Even I tried what Simon said with a simulator called GEOGEBRA. I get the same result.
 
I get the same result as you using a + b + c = 180° = (180 - b)/2 + b + a + c/2.
 
I guess there must be a printing error in the question like may be the angles changed or like that
 
Simon Bridge said:
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.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
5
Views
3K
Replies
4
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K