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In $\triangle{ABC}$, $\angle{ABC}=20^{\circ}$, $\angle{ACB}=30^{\circ}$, $D$ is a point inside the triangle and $\angle{ABD}=10^{\circ}$, $\angle{ACD}=10^{\circ}$, find $\angle{CAD}$.

Note: I have seen some very similar question with beautiful solution in pure geometric format. I know how to solve this problem in trigonometric format. But I think this problem deserves a beautiful geometric approach as solution, and that's why I post it here.

As request, here is approach applying Ceva's theorem in trigonometric form,

$$\begin{align*} \frac{\sin130}{\cos130+2\cos10}&=\tan(x)\Longrightarrow \frac{\sin120\cos10+\cos120\sin10}{\cos120\cos10-\sin120\sin10+2\cos10}\\ &=\frac{\sqrt{3}\cos10-\sin10}{3\cos10-\sqrt{3}\sin10}.\\ &=\frac{1}{\sqrt{3}}\\ &=\tan30\Longrightarrow x\\ &=\boxed{30}\\ \end{align*}$$

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Jessie
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r ne
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    You should know that the community prefers/expects a question to include something of what the asker knows about the problem. (What have you tried? Where did you get stuck? etc) This helps answerers tailor their responses to best serve you, without wasting time (theirs or yours) explaining things you already understand or using techniques beyond your skill level. (It also helps convince people that you aren't simply trying to get them to do your homework for you. An isolated problem statement with no evidence of personal effort makes a poor impression, attracting down- and close-votes.) – Blue Apr 22 '21 at 09:16
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    I can help but before that, you need to show your effort. Did you at least attempt using Trigonometric form of Ceva's theorem or law of sines? Did you get success? If you tried a geometric solution, what construction did you do? Where did you get stuck? – Math Lover Apr 22 '21 at 09:27
  • @MathLover I can easily solve it using trigonometric knowledge, analytics. This question is to ask for a pure geometric solution which means middle school students level. – r ne Apr 22 '21 at 09:41
  • @r ne: FADYPaW. For assistance do (demonstrate) your part of work. Did you at least chase/label the angles around D? – Narasimham Apr 22 '21 at 09:46
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    The point is not whether you can solve it. Did you solve it? If yes, then why not share all your work and turn this a good question as per site guidelines? – Math Lover Apr 22 '21 at 09:48
  • @Blue I have seen similar question like https://apfstatic.s3.ap-south-1.amazonaws.com/s3fs-public/15-comac_20-30-130_addendum.pdf and it contains a good solution. I tried with several ways like mirroring triangle ACD along AD, it's like you know it's forming equilateral triangle and very close but not getting to it. Actually I was working on another similar question, with $\angle{ABC}=80^{\circ}, \angle{ABP}=20^{\circ}, \angle{ACB}=50^{\circ}, \angle{ACP}=10^{\circ}$. I formed several parallel lines and it's also close but not getting to the result. – r ne Apr 22 '21 at 09:48
  • @Narasimham No, simple angle chasing is not easy for this one. – r ne Apr 22 '21 at 09:50
  • Right. so I said at least... If you indicated part of your response to @Blue I suppose that counts as due deligence. – Narasimham Apr 22 '21 at 09:52
  • @rne: That's all good information that should be included in your question. (Comments are easily overlooked and may be hidden.) In particular, you should mention that you have already seen a geometric solution in that PDF (summarizing it in the body of the question would be helpful), and you should indicate what you find unsatisfactory about it. Otherwise, answerers risk spending their time duplicating that solution or providing another that you'll find unsatisfactory. ... When you're asking someone for help, you want to do all you can to ensure that the someone's efforts aren't wasted. – Blue Apr 22 '21 at 09:58
  • @r ne: Right. so I said at least... If you indicated part of your response to @Blue in your question I suppose that counts as due deligence. – Narasimham Apr 22 '21 at 09:59
  • @MathLover Okay, here is what i did after i saw your last post: applying Ceva's theorem in trigonometric form, $\dfrac{sin130}{cos130+2cos10}=tan(x) \implies \dfrac{sin120cos10+cos120sin10}{cos120cos10-sin120sin10+2cos10}=\dfrac{\sqrt{3}cos10-sin10}{3cos10-\sqrt{3}sin10}=\dfrac{1}{\sqrt{3}}=tan30 \implies x=\boxed{30}$. This is trivial. The question here asks for a pure geometric solution. – r ne Apr 22 '21 at 10:00
  • @Blue That PDF is for a different problem. It's perfect but not the same problem. – r ne Apr 22 '21 at 10:02
  • @rne: Whoops. If I'm going to criticize someone's writing, I should do a better job of reading. :) ... Still, I'll reiterate: Add context and clarifications and whatnot into the body of the question. (Especially with so much "discussion" going-on in the comment section, there's a good chance that all the information you've provided will get migrated to a chat where no one will see it (and formatting will be lost).) – Blue Apr 22 '21 at 10:16
  • @Blue I think posting a geometric problem and ask if anyone has a pure geometric solution for it, is already a perfect form of question. At least if I see someone's question like this, I have fully understood what it means. There are always lots of similar questions and knowledge around to "show efforts" but why not take them as optional hints. – r ne Apr 22 '21 at 10:21
  • @rne: I'm not trying to argue, but ... An isolated problem statement provides no way to tell that you're personally invested in a solution; for all the reader knows, it might well be a do-my-homework-for-me problem. "Showing efforts" gives a question its best chance of being taken seriously. But also, showing trig work (even just giving the answer), linking to that other problem, describing what you've considered, etc, may spark insights to someone else, while saving them time in duplicating that effort. It's easier to forge a "better" path when you've seen where others have tread. Cheers! – Blue Apr 22 '21 at 10:34
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    @Blue Yeah I know what you mean now. You are right that people would not like a do-my-homework-for-me question. That's a good reminder question posting, truly. I have added a note to the question. Last time I posted a geometric problem and provided my algebraic solution and asked for pure geometric approach, the admin saw it and thought I was showing off, and closed my question. So it's hard to guess what people think and I am still learning on that part... – r ne Apr 22 '21 at 10:44
  • @Blue Maybe next time before asking a question, i should add: my favorite bands are Beatles and ABBA, and I love Maradona. I have been playing Go the board game for more than 20 years and you would not think this old man are still working on his middle school homework asking help for a pure geometric solution on this problem, here it is:... – r ne Apr 22 '21 at 10:56
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    @rne ok based on your comment with trigonometric solution and additional context in the question, I am posting a geometric solution in sometime. I would still request if you could edit your post with trigonometric solution that you put in comments. Nobody (at least I) likes to spend time answering a question that would later get closed or deleted. Please see from answerer's point of view too. – Math Lover Apr 22 '21 at 11:05
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    @MathLover Done. Added. – r ne Apr 22 '21 at 11:12

2 Answers2

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Please extend line segment $BA$. We have $\angle CAE = 50^0$. Draw $\angle ACE = 50^0$. We have $CE = AE$.

So, $\angle BCE = \angle BEC = 80^0$. $BM$ is angle bisector of isosceles triangle $\triangle CBE$ where $BC=BE$.

Therefore $CD = DE$. As $\angle DCE = 60^0$, $\triangle DCE$ is equilateral triangle and $DE = CE = AE$. So $\triangle AED$ is isosceles triangle with $\angle AED = 20^0$.

That leads to $\angle DAE = 80^0 \implies \angle DAC = 30^0$.

Math Lover
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COMMENT.-This could be another nice way to prove that $x=30º$.

enter image description here

Piquito
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  • Here you have A, E, G determined, so F is also determined, now you have to prove that $\triangle{EFC}$ is isosceles. If this is proven, the rest is correct. – r ne Apr 26 '21 at 03:30