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 $ABC.$

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triangle BDI is a right triangle (sides are 3-4-5)

That means that triangle ABC is isosceles, since the angle bisector AD is also an altitude.
That means that
BC = 2*BD = 8
AD is also a median of ABC, so AI = 2*DI = 6
The area of ABC is thus 6*8/2 = 24

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