# Need a pure geometric solution to a $20-30-130$ triangle question

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*} • 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 at 9:16
• 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 at 9:27
• 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 at 9:48
• @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 at 10:44
• @MathLover Done. Added. – r ne Apr 22 at 11:12 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 $$\angle DAE = 80^0 \implies \angle DAC = 30^0$$.

• Cool. This is very like the PDF solution. I should have got this. Thank you! – r ne Apr 22 at 11:37
• you are welcome. – Math Lover Apr 22 at 11:38

COMMENT.-This could be another nice way to prove that $$x=30º$$. • 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 at 3:30