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This identity is solvable by the help of trigonometry identities, but I guess there is an interesting and simple geometry interpretation behind this identity and I can't find it.

I found it when I was thinking about World's Hardest Easy Geometry Problem

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  • $\begingroup$ Can anyone solve it via trigonometric identities? $\endgroup$ – zz20s Jan 24 '16 at 18:41
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    $\begingroup$ The diagrams I posted give a geometric interpretation of the claim, but not a proof: I mistook a sum-of-angles condition for a constraint. Consequently my proposed argument doesn't use the specific angle values, and can't be fixed without some "genuinely new" ingredient. $\endgroup$ – Andrew D. Hwang Jan 24 '16 at 19:54
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    $\begingroup$ @zz20s one way of proving it using non-geometrical means is to consider the equation $\tan 5x+\tan4x =0$ both as a simple trig equation whose roots are integer multiples of $20$ degrees and as a polynomial in $t=\tan x$, whose product of roots will give the result... $\endgroup$ – David Quinn Jan 24 '16 at 20:02
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Consider a triangle $ABC$ with $\angle A = 10^\circ$, $\angle B = 150^\circ$, and $\angle C=20^\circ$. Let $O$ be the circumcenter of the triangle $ABC$. Then $$\angle AOC = 360^\circ - 2 \angle B = 60^\circ,$$ so triangle $AOC$ is equilateral.

Let $S$ be the circumcenter of the triangle $OBC$. Then $$\angle BSC = 2\angle BOC = 4\angle BAC = 40^\circ$$ and $$\angle SCB = \frac{180^\circ - \angle BSC}2 = 70^\circ,$$ so $\angle SCA = 50^\circ$. Denoting the intersection of $AC$ and $SB$ by $X$ we get $$\angle CXS = 180^\circ - \angle SCA - \angle BSC = 90^\circ$$ so $AC \perp SB$.

Moreover $\triangle ASC \equiv \triangle ASO$ because $AC=AO$ and $SC=SO$. In particular $\angle CAS = \angle SAO$ and since $\angle CAO=60^\circ$, we have $\angle CAS = 30^\circ$.

Observe that $$\frac{\tan \angle BAC}{\tan \angle ACB} = \frac{\frac{BX}{AX}}{\frac{BX}{CX}} = \frac{CX}{AX} = \frac{\frac{SX}{AX}}{\frac{SX}{CX}} = \frac{\tan \angle CAS}{\tan \angle SCA},$$ therefore $$\frac{\tan 10^\circ}{\tan 20^\circ} = \frac{\tan 30^\circ}{\tan 50^\circ}.$$

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