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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
 #1
avatar+824 
+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 u2

 

 Jan 4, 2021
edited by jugoslav  Jan 4, 2021
 #2
avatar+428 
+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 :)

 Jan 4, 2021
edited by Guest  Jan 4, 2021
edited by Pangolin14  Jan 4, 2021
 #3
avatar
+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...

 Jan 4, 2021
 #5
avatar+112000 
+1

Pangolin,

You have not explained how you got the point A.

Remember, no trig is allowed.

Melody  Jan 5, 2021
 #6
avatar+428 
+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
edited by Pangolin14  Jan 5, 2021
 #7
avatar+824 
+1

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

jugoslav  Jan 5, 2021
 #8
avatar+112000 
+2

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

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

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

 

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

 

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

 

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

 

 

 

 

 Jan 6, 2021
edited by Melody  Jan 6, 2021
 #9
avatar+428 
+2

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

 Jan 8, 2021
 #10
avatar+112000 
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.  wink

Melody  Jan 8, 2021

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