# 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.

• BDI actually isn't similar to BFI. I have an answer, but it requires trigonometry, and that seems like it would be out of scope. Jun 18, 2021 at 23:51
• @DougM - I can also solve it with double angle formulae from trigonometry to get a rational answer, but do not know whether that is allowed Jun 19, 2021 at 0:01
• Probably not, I'm taking an intro to geometry class, trig comes in the next class, so not allowed I believe. Jun 19, 2021 at 0:02

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 $$0.$$

$$AI = \frac {75}{7}\\ AD = \frac {75}{7} + 3 = \frac {96}{7}$$

$$Area = \frac 12(AC)(AD) = 4 \frac {96}{7} = \frac {384}{7}$$

• When you factor the quadratic into $(7x+25)(x-1)=0$, shouldn't it be $(7x-25)(x+1)=0$? Jun 19, 2021 at 0:56
• @DanielGeyfman thanks! Jun 21, 2021 at 17:06

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 $$2 \times \frac12 \times\frac{96}{7}\times 4= \frac{384}{7}\approx 54.857$$.

If $$AD\bot BC\Rightarrow \Delta ABC:AB=AC$$ (isosceles triangle where onl one angle bisector is also a height).

The triangle can’t be equilateral because in that case $$BI=AI=2ID\Leftrightarrow 5=6$$

Now to calculate the area:

$$\begin{array}{lrlr}area(\Delta ABC)&=&area(\Delta BIC)+2\cdot area(\Delta AIB)\\&=&2\cdot area(\Delta BDI)+2\cdot(area(\Delta BHI)+area(\Delta AHI), IH\bot AB\end{array}$$

Then $$\Delta BDI \equiv \Delta BHI\Rightarrow area(\Delta BHI)=area(\Delta BDI)$$ $$\Delta BDI \sim \Delta IHA\Rightarrow area(\Delta AHI)=area( \Delta BDI)\cdot \left(\frac{BD}{HI}\right)^2$$

I hope this is enough to get you going. If you need more help let me know.