Let ABC be a triangle, and let its angle bisectors be AD, BE, and CF which intersect at I. If DI=3, BD=4 and BI=6 then compute the area of triangle BID.

omgitsrne Feb 7, 2024

#1**+1 **

Here is a solution that doesn't require any complex geometry. We see that we know all 3 side lengths of triangle BID. We know a formula that can tell us the area directly knowing 3 sides of the triangle, Heron's formula.

First we calculte the semi-perimeter \(\frac{3+4+6}{2}=\frac{13}{2}\).

Then we apply Heron's formula. \(Area = \sqrt{\frac{13}{2}(\frac{13}{2}-3)(\frac{13}{2}-4)(\frac{13}{2}-6)}\). Simplifying, we get \(Area = \sqrt{\frac{455}{16}}=\frac{\sqrt{455}}{4}\)

So, the Area is **sqrt(455)/4** which is around 5.3327.

hairyberry Feb 9, 2024