In $\triangle ABC, AB = 12, AC = 10$.
$I$ is incenter $∠BIC = 105 ^{\circ}$.
Find area of $\triangle ABD$ where $AD$ is angle bisector.

I've drawn the following figure: Figure

Now, $∠IBD + ∠ICB =75 ^{\circ} $
Hence $∠A=180-150=30^{\circ} $
By law of cosines ,
Now , $x+y=12$

Now I want the length of perpendicular from $A$ to $BC$ . The first problem is that I can't find it.
The second problem is that I think my algebraic approach is not at all elegant , so can anyone find a better way to solve this problem?

  • $\begingroup$ You can use ^{\circ} for degree instead of deg. $\endgroup$ – Sawarnik Apr 3 '14 at 8:54
  • $\begingroup$ @AGoogler (1) In your work, if $BC^2=244-120\sqrt3$, then $y+z=244-120\sqrt3$ is not true. (2) The method used by Ajay is the correct way of solving it. The approach he is using is called "bisector theorem". $\endgroup$ – Mick Apr 3 '14 at 9:27
  • $\begingroup$ @mick Yes , it should be the square root of that. And yes I know about that theorem but didn't use it. $\endgroup$ – A Googler Apr 3 '14 at 10:00
  • $\begingroup$ @Mick Why do you think my method is not correct? I think its very standard and general, in addition gives you the length of the angle bisector easily. $\endgroup$ – Sawarnik Apr 3 '14 at 14:14
  • $\begingroup$ @Sawarnik By agreeing Ajay's method is correct only means I would have the problem solved the same way. It by no mean of implying other's (including yours) is incorrect. After re-viewing your work, I also agree that it is "very general and standard". Sorry for creating such mis-understanding. $\endgroup$ – Mick Apr 3 '14 at 17:22

I don't know whether its invertendo - compenendo - invertendo or something else. These answers were intended for me so they are written in the language that I can understand. So if you didn't got something do ask.


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  • $\begingroup$ Thanks Ajay , your approach is nice too. I like the way you solve the questions - with question in red ink , explanation for each step etc. It'd so much better if I were like you. I just solve problems with pencil and like doing rough work. How do you manage to be so neat? $\endgroup$ – A Googler Apr 3 '14 at 9:07
  • $\begingroup$ @AGoogler I was thinking the same thing too! Its so neat! I am completely opposite however, I do things in the most haphazard manner and don't write anything unless it isn't obvious. $\endgroup$ – Sawarnik Apr 3 '14 at 9:15
  • $\begingroup$ I wrote all answers of MPA for future reference. I am the most clumsiest person you would ever find. I just sit anywhere making legs touching walls up, use anything I find for writing even a chalk on floor. So don't be sad It's just for MPA. $\endgroup$ – Ajay Apr 3 '14 at 16:05
  • $\begingroup$ I only write main calculations while solving questions for speed. Accuracy comes by experience. If I have to present it to someone then only I write like this. $\endgroup$ – Ajay Apr 3 '14 at 16:09
  • $\begingroup$ @Ajay Oh, then your presentation has got to be the clearest I have ever seen. $\endgroup$ – Sawarnik Apr 3 '14 at 16:13

We know $\frac{\angle A}{2}=15^{\circ}$.

Now by summing the two areas $(ABD)$ and $(ACD)$ we get $(ABC)$, using the area formula $2\Delta=ac\sin B$, we can write it as: $$AD\cdot12\sin15^{\circ}+AD\cdot10\sin 15^{\circ}=120\sin 30^{\circ}$$ So, $$AD=\frac{30}{11\sin15^{\circ}}$$ Thus, $(ABD)=\frac12\cdot12\cdot AD \cdot\sin 15^{\circ}= 6\cdot\frac{30}{11\sin15^{\circ}}\cdot\sin 15^{\circ}=\frac{180}{11}$.

  • $\begingroup$ Can you please explain your answer? Which two areas are you summing? $\endgroup$ – A Googler Apr 2 '14 at 15:28
  • 1
    $\begingroup$ @AGoogler ABD and ACD. Isnt it obvious? We are using $2A=bc\sin A$. $\endgroup$ – Sawarnik Apr 2 '14 at 15:30
  • $\begingroup$ Okay got it, Thanks. $\endgroup$ – A Googler Apr 2 '14 at 15:32
  • $\begingroup$ @AGoogler Ok :) Is my final answer correct? $\endgroup$ – Sawarnik Apr 2 '14 at 15:34
  • 1
    $\begingroup$ Yes , It is correct :) $\endgroup$ – A Googler Apr 2 '14 at 15:37

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