In triangle ABC, angle bisectors AD,CF and BE meet at I. If DI=3, BD=4, and BI=5, then compute the area of triangle ABC.

I know many people already answered this question, but as a challenge, can you guys solve it another way?

If you want to see what I did so far:

After drawing the diagram, we can start by finding congruent triangles. We know that $\triangle{BDI}$ is a right triangle because it is a $3-4-5$ triangle. From here we know that $\angle BDI = 90^\circ \mbox{ and } \angle {CDI} = 90^\circ.$ Since $\overline{AD}$ is the angle bisector, $\angle{DAC} \cong \angle{DAB}$. We also obviously know that overlineAD congruent overline{AD}. Finally, since angle BDI congruent angle CDI = 90 degrees we know that triangle BAD congruent triangle CAD by ASA congruence. From this congruence we know that overline{DC} = 4 and angle ABD congruent angle ACD by CPCTC. We can now safely say that triangle BID congruent triangle CID by SSS congruence since both triangles are 3-4-5 triangles. We then know that overline{IC} = 5 by CPCTC

Guest Jan 4, 2021

#1**+2 **

In a triangle, ABC, angle bisectors AD, CF and BE meet at a point I. If DI=3, BD=4, and BI=5, then compute the area of triangle ABC.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

∠B = 2(tan^{-1}(3/4)) = 73.73979529º

AD = tan∠B * 4 = 13.71428571

Area = BD * AD = 54.857142286 u^{2}

jugoslav Jan 4, 2021

#2**+2 **

You are off to a good start. We know that because BDI is a 3-4-5 triangle, it is right, and because an angle bisector is a perpendicular in triangle ABC, ABC is isosceles. As ABC is isosceles, a perpendicular from A is a median of BC! Therefore, BC = 2*BD = 8.

I'm going to coordinate bash. Let B be (-4,0) and C be (4,0). Then, reverse-using the formula for the incenter's coordinates, or (0,3), A = (0, 96/7).

We have that the height is 96/7, so the area is 96/7 * 4 = 54.8571428571 or 54 6/7.

jugoslav, nice to see we agree :)

Pangolin14 Jan 4, 2021

#3**+2 **

Can you solve it another way? Or can you continue with the work I have? I haven't learned about tan and stuff yet so...

Guest Jan 4, 2021

#5**+1 **

Pangolin,

You have not explained how you got the point A.

Remember, no trig is allowed.

Melody
Jan 5, 2021

#6**+2 **

I didn't use trig- I'll explain.

The incenter coordinates are $\left(\frac{1}{2} (-a + b + c), \frac{\sqrt{-(a - b - c) (a + b - c) (a - b + c) (a + b + c)}}{{2 (a + b + c)}} \right)$ where a,b,and c are the side lengths of a triangle. Then we have that the incenter's coordinates, if we let a=8 and b = c, are $\left(\frac{1}{2} (-8 + b + b), \frac{\sqrt{-(8 - b - b) (8 + b - b) (8 - b + b) (8 + b + b)}}{2 (8 + b + b)}\right)$ which simplifies to $\left(\frac{1}{2} (2 b - 8), 2 \frac{\sqrt{(2 b - 8) (2 b + 8)}}{ b + 4}\right).$ If we set the right side of the equation equal to 3, we can easily cross multiply and get $b = \frac{100}{7}.$ Thus, AB = $\frac{100}{7},$ and we may want to use heron's formula, but for a more convenient solution we can say that A = (0,a), and using the distance formula with B = (-4,0), we have $\sqrt{a^2-16} = 100/7,$ so $a = \frac{96}{7},$ and the height is $\frac{96}{7}$

Pangolin14
Jan 5, 2021

#8**+2 **

Thanks Pangolin14 and Yugoslav

I tackled this in a different way.

I also used co-ordinate geometry. As shown on my pic.

I set the centre of the incircle (since that is what I is) as the origin.

B(-4,-3) and C((4,-3)

I need to find the height of the triangle

Let the equation of the line AB be y= -mx+b where m and b are both positive.

The perpendicular distance of the origin to line AB is the radius, 3.

So I will use the perpendicular distance formula

y= -mx+b

mx + y -b = 0 (0,0)

\(3=\frac{|0+0-b|}{\sqrt{m^2+1}}\\ 3=\frac{b}{\sqrt{m^2+1}}\\ b=3\sqrt{m^2+1} \)

So we have

\(y=-mx+3\sqrt{m^2+1}\\\)

This line passes through the point (4,-3) so substitute

\(-3=-4m+3\sqrt{m^2+1}\\ -3+4m=3\sqrt{m^2+1}\\ 16m^2-24m+9=9(m^2+1)\\ 16m^2-24m+9=9m^2+9\\ 7m^2-24m=0\\ 7m-24=0\\ m=\frac{24}{7}\)

\(b=3\sqrt{m^2+1}\\ b=3\sqrt{(\frac{24}{7})^2+1}\\ b=3*\frac{25}{7}=\frac{75}{7}\\ \text{height of triangle }= \frac{75}{7}+3 = \frac{96}{7}\)

\(\text{Area of triangle ABC = }\frac{1}{2}*8*\frac{96}{7}\\ \text{Area of triangle ABC = }54\frac{6}{7}\;units^2\)

Melody Jan 6, 2021

#9**+2 **

That's an elegant solution, Melody. Definitely better than bashing with incircle coordinates

Pangolin14 Jan 8, 2021

#10**0 **

**That is really nice of you to say Pangolin. Your solution is impressive as well.**

Both of our solutions require a lot of geometry knowledge. Especially if you were doing it under test conditions and could not look things

up.

For mine, and I think yours as well, the answerer would have to know that the point I is the centre of an incircle.

For mine, an answerer needs to know the perpendicular distance formula.

This was an interesting question and I am glad I was pushed to answer it without using trigonometry.

It would of course be nice to get proper feedback from the asker.

Melody
Jan 8, 2021