Question

In triangle $abc,$ angle bisectors $\overline{ad},$ $\overline{be},$ and $\overline{cf}$ meet at $i.$ if $di = 3,$ $bd = 4,$ and $bi = 5,$ then compute the area of triangle $abc.$

Answer 1

∆BDI is necessarily a right triangle, since the squares of the two smaller sides is exactly the square of the longest side.

3² + 4² = 9 + 16 = 25 = 5²

Then the angle bisector AD forms a right angle with the base BC - this means AD is an altitude of ∆ABC.

Since AD bisects the angle at vertex A, we further know that ∆ABC is isosceles, since both ∆ABD and ∆ACD are right triangles with the same height AD. Then the length of BC is twice the length of BD, and ∆ABC has a base of length 8.

Let θ be the measure of one of the base angles of ∆ABC (angle ABC, for instance). Then in ∆BDI, we have the trigonometric relationship

tan(θ/2) = 3/4

from which we get

θ/2 = arctan(3/4)

θ = 2 arctan(3/4) ≈ 73.74°

Then the height of ∆ABC is

tan(θ) = AD/4

AD = 4 tan(θ)

AD = 4 tan(2 arctan(3/4))

AD = 4 • 2 tan(arctan(3/4)) / (1 - tan²(arctan(3/4)))

AD = 4 • 2 • 3/4 / (1 - 9/16) = 96/7

Then the area of ∆ABC is

1/2 • 8 • 96/7 = 384/7