+400 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.
+399 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.
+1632 Nice! Heron's Formula is a very important formula: The area of a triangle with side lengths \(a\), \(b\), and \(c\), with semi-perimeter = \(a + b + c\over{2}\) is:
\(\sqrt{s(s-a)(s-b)(s-c)}\)
This formula will come in handy, especially if you are doing competition math. 
