# Thread: Special Segments in Triangles

1. ## Special Segments in Triangles

This was on a special segments sheet, I know the answer is 9, but I don't know how to prove it.

2. ## Re: Special Segments in Triangles

Call the point where the vertical line intersects BC point O. Assuming the vertical line is perpendicular to BC (is it?), then angle BOA = COA = 90 degrees. Then you have triangles ABO and ACO are equivalent from side-angle-side. Hence AB = AC.

3. ## Re: Special Segments in Triangles

Originally Posted by kamnarra

This was on a special segments sheet, I know the answer is 9, but I don't know how to prove it.
I see no reason that it should equal 9. I see no reason to assume there are any right angle(s) in the figure.

4. ## Re: Special Segments in Triangles

Sorry, there was a right angle. I forgot to draw it.

5. ## Re: Special Segments in Triangles

Originally Posted by kamnarra
Sorry, there was a right angle. I forgot to draw it.
Well then, next time please review before posting making sure all relevant information is given.
The only triangle in which an altitude is a bisector of its base is an isosceles triangle.