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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.

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  • $\begingroup$ 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. $\endgroup$
    – Doug M
    Jun 18, 2021 at 23:51
  • $\begingroup$ @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 $\endgroup$
    – Henry
    Jun 19, 2021 at 0:01
  • $\begingroup$ Probably not, I'm taking an intro to geometry class, trig comes in the next class, so not allowed I believe. $\endgroup$
    – dgeyfman
    Jun 19, 2021 at 0:02

3 Answers 3

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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.

enter image description here

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 $0.$

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

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

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  • $\begingroup$ When you factor the quadratic into $(7x+25)(x-1)=0$, shouldn't it be $(7x-25)(x+1)=0$? $\endgroup$
    – dgeyfman
    Jun 19, 2021 at 0:56
  • $\begingroup$ @DanielGeyfman thanks! $\endgroup$
    – Doug M
    Jun 21, 2021 at 17:06
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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 $2 \times \frac12 \times\frac{96}{7}\times 4= \frac{384}{7}\approx 54.857$.

enter image description here

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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.

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