Hopefully CPhill or Melody sees this and helps me out.
I don't want/need the full solution to this problem, as I just need help figuring out the problem.
Any hints would be fine.
Here's the problem:
In triangle ABC, angle bisectors AD,CF and BE meet at I. If DI=3, BD=4, and BI=5, then compute the area of triangle ABC.
Please help me out! Any help would be greatly appreciated
In triangle ABC, angle bisectors AD,CF and BE meet at I. If DI=3, BD=4, and BI=5, then compute the area of triangle ABC.
Hello EinsteinBrain29!
To solve your question, I need the following information:
Point D is the intersection of the bisector AD with the side BC,
or
point D is the point of contact of the inner circle with side BC? Then ID would be the radius of the inner circle.
I ask for a information.
!
In a triangle, ABC, angle bisectors AD, CF and BE meet at a point I. If DI=3, BD=4, and BI=5, then compute the area of triangle ABC.
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The following is what we know:
DI = 3 BD = 4 BI = 5
FI = EI = DI = 3
∠AEB ≅ ∠CEB = 90º ∠IBD = tan-1(3 / 4)
This problem can be easily solved by using this information.
Have fun.
Oh thanks for the hints! I think I can try to take it on from here! I'll reply back in here when I bump into problems.