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

 Jan 4, 2021

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 u2


 Jan 4, 2021
edited by jugoslav  Jan 4, 2021

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 :)

 Jan 4, 2021
edited by Guest  Jan 4, 2021
edited by Pangolin14  Jan 4, 2021

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...

 Jan 4, 2021


You have not explained how you got the point A.

Remember, no trig is allowed.

Melody  Jan 5, 2021

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
edited by Pangolin14  Jan 5, 2021

Great job, Pangolin! I was thinking about this problem but I couldn't come up with non-trig solutions.

jugoslav  Jan 5, 2021

Thanks Pangolin14  and Yugoslav     laugh


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


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


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


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


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





 Jan 6, 2021
edited by Melody  Jan 6, 2021

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

 Jan 8, 2021

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


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.  wink

Melody  Jan 8, 2021

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