Angle bisectors of triangles 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 $ABC.$
We are learning about angle/perpendicular bisectors in my geometry class, but I don't fully understand them. I figured out that $\triangle BDI$ must be right by Pythagorean Theorem, and that $\triangle BDI \cong \triangle BFI$ by SAS, but I don't know how to go from here. I feel like it would have something to do with the Angle Bisector Theorem, which we learned, but I don't know for sure.
EDIT: I assigned $x=CI$ and $y=AI$, and solved for every side length in terms of $x$, $y$, and the values I already solved for, but I still don't know how to continue.
 A: Without using trig.
From the fact that $\triangle BDI$ has side lengths $3,4,5$ and the Pythagorean theorem, we know that angle D is a right triangle.
If $D$ is a right angle, $\triangle ABD$ is isosceles.

This figure is only half of ABC...
There is a theorem regarding bisected angles that says that if $IB$ bisects $B$ then $AI:AB$ as $DI:DB$
$AI: AB =  3:4$
Let $AI = 3x, AB = 4x$
From Pythagoras:
$BD^2 + (DA)^2 = AB^2\\
4^2 + (3 + 3x)^2 = (4x)^2$
Now for some algebra:
$16 + 9 + 18x + 9x^2 = 16x^2\\
7x^2 - 18x - 25 = 0\\
(7x - 25)(x + 1)$
We can reject, $x = -1$ as lengths must be greater than $0.$
$AI = \frac {75}{7}\\
AD = \frac {75}{7} + 3 = \frac {96}{7}$
$Area = \frac 12(AC)(AD) = 4 \frac {96}{7} = \frac {384}{7}$
A: As a slight alternative to Doug M's answer,
if $\angle BDI$ is a right angle then so is $\angle CDI$ so you have symmetry across $AD$ and an isosceles triangle - which helps a lot.  You also have $DI$ as the radius of the in-circle, and if you join $I$ to the other two points $G$ and $H$ where the edges touch the in-circle then you do get similar triangles $\triangle ADB$ and $\triangle AGI$, as well as congruent triangles $\triangle IDB$ and $\triangle IGB$.
So $\dfrac{AG}{3} = \dfrac{AD}{4}$ and $(AG+4)^2 = AD^2+4^2$
which you can solve to give $AD=\frac{96}{7}$
and so the area is $2 \times \frac12 \times\frac{96}{7}\times 4= \frac{384}{7}\approx 54.857$.

