# Angle bisector divides the triangle into two triangles. Find the area of one of them.

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: Now, $$∠IBD + ∠ICB =75 ^{\circ}$$
Hence $$∠A=180-150=30^{\circ}$$
By law of cosines ,
$$BC^2=244-120\sqrt3$$
Now , $$x+y=12$$
$$x+z=10$$
$$y+z=\sqrt{244-120\sqrt3}$$

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?

• You can use ^{\circ} for degree instead of deg. – Sawarnik Apr 3 '14 at 8:54
• @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". – Mick Apr 3 '14 at 9:27
• @mick Yes , it should be the square root of that. And yes I know about that theorem but didn't use it. – A Googler Apr 3 '14 at 10:00
• @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. – Sawarnik Apr 3 '14 at 14:14
• @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. – Mick Apr 3 '14 at 17:22

## 2 Answers

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.  • 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? – A Googler Apr 3 '14 at 9:07
• @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. – Sawarnik Apr 3 '14 at 9:15
• 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. – Ajay Apr 3 '14 at 16:05
• 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. – Ajay Apr 3 '14 at 16:09
• @Ajay Oh, then your presentation has got to be the clearest I have ever seen. – 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, $(ABD)=\frac12\cdot12\cdot AD \cdot\sin 15^{\circ}= 6\cdot\frac{30}{11\sin15^{\circ}}\cdot\sin 15^{\circ}=\frac{180}{11}$.

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