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In $\triangle ABC$ with $AB=AC$ and $\angle BAC=20^\circ$, point $D$ is on $AC$, with $BC=AD$. Find $\angle DBC$.

I know the correct solution, but I'm more interested in where is the problem in my solution.

My solution :

Geo figure

Now in $\triangle ABD$, applying the sine rule:

$$\frac{AD}{\sin\alpha} = \frac{BD}{\sin 20^\circ} \tag{1}$$

In $\triangle BDC$:

$$\frac{BD}{\sin 80^\circ} = \frac{BC}{\sin(180^\circ-\beta)} \tag{2}$$

We know $AD= BC$; put in to $(1)$:

$$\frac{BC}{\sin\alpha} = \frac{BD}{\sin 20^\circ} \tag{3}$$

Comparing $(2)$ and $(3)$:

$$\frac{BC}{BD} = \frac{\sin\alpha}{\sin 20^\circ} = \frac{\sin(180^\circ-\beta)}{\sin 80^\circ} \tag{4}$$

$$\frac{\sin \alpha}{\sin(180^\circ-\beta)} = \frac{\sin 20^\circ}{\sin 80^\circ} \tag{5}$$

Now, $\alpha = 20^\circ$ and $\beta = 100^\circ$, but when I plug these values in $\triangle ABC$, it's not even triangle. oO

Where I am wrong? Thanks.

PS : sorry for poor editing, I don't have any clue about it.

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    $\begingroup$ Full marks for the diagram, but please visit the MathJax documentation page and shore up the above by yourself. Note that MathJax is much easier to read, and a question that is good to read gets more attention, so fifteen minutes of learning the basics gets you a lot of attention on your questions and better answers. $\endgroup$ – астон вілла олоф мэллбэрг Jan 19 at 9:27
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    $\begingroup$ @AngelusMortis: I'm making some edits to show you how it's done. Please don't edit until I'm finished. :) $\endgroup$ – Blue Jan 19 at 9:31
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    $\begingroup$ Good to see you taking the message on board. As for the actual question, from $\frac{\sin(\alpha)}{\sin(180-\beta)} = \frac{\sin 20}{\sin 80}$, how did you get to $\alpha = 20$ and $\beta = 100$? That part you have not explained. Ok, I see : you just compared numerators and denominators, and just took $\alpha = 20$ and $180-\beta = 80$ so $\beta = 100$. That must be the incorrect step here, then. $\endgroup$ – астон вілла олоф мэллбэрг Jan 19 at 9:31
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    $\begingroup$ I see. But that does not work here, unfortunately. Instead, note that $\sin (180 - \beta) = \sin \beta$ from the sum of sines formula, and also that $\beta + \alpha = 160$ so $160 - \alpha = \beta$. So substituting, gets you $\frac{\sin \alpha}{\sin(160-\alpha)} = \frac{\sin 20}{\sin 80}$. See what you can do from here. Also, since you know the correct value of $\alpha$, check that it satisfies the above equation. $\endgroup$ – астон вілла олоф мэллбэрг Jan 19 at 9:37
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    $\begingroup$ @Blue Thanks for helping with editing. Much appreciation sir :) $\endgroup$ – Angelus Mortis Jan 19 at 10:59
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So we have that $\frac{\sin \alpha}{\sin (180-\beta)} = \frac{\sin 20}{\sin 80}$.

The first thing we use is that $\alpha + \beta = 160$ from the triangle $ABD$. From here, $180 - \beta = 180 - (160-\alpha) = 20 + \alpha$.

Next, we note that: $$ \frac{\sin 20}{\sin 80} = \frac{\sin 20}{\cos(90-80)} = \frac{\sin 20}{\cos 10} = \frac{2 \sin 10 \cos 10} {\cos 10} = 2 \sin 10 $$

So, we have the equation : $$ \frac{\sin \alpha}{\sin (\alpha + 20)} = 2 \sin 10\\ \implies \sin \alpha = 2 \sin 10 \sin (20+\alpha) = 2 \sin 10 \sin 20 \cos \alpha + 2 \sin 10 \cos 20 \sin \alpha $$

Now, collecting terms of $\sin \alpha$ on one side, and facctorizing it out, $$ \sin \alpha(1 - 2 \sin 10 \cos 20) = 2 \sin 10 \sin 20 \cos \alpha \\ \implies \tan \alpha = \frac{2 \sin 10 \sin 20}{1 - 2 \sin 10 \cos 20} $$

The right hand side is some fixed number which we have to find.

To do this, we first simplify the denominator, using the formulas : $$2 \sin A\cos B = \sin(A+B) + \sin(A-B) \quad ; \quad \sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$$

We will also use the fact that $\sin 30 = \frac 12$. In our case, $$ 1 - 2 \sin 10 \cos 20 = 1- (\sin 30 + \sin (-10)) = 2 \sin 30 - (\sin 30 - \sin 10) \\ = \sin 30 +\sin 10 = 2 \sin 20 \cos 10 $$

Therefore, $$ \tan \alpha = \frac{2 \sin 10 \sin 20}{1 - 2 \sin 10 \cos 20} = \frac{2\sin 10 \sin 20}{2 \cos 10 \sin 20} = \tan 10 $$

Now, since $0 < \alpha < 180$, we get that $\alpha = 10$. From here, $80-\alpha = 70$ is the desired angle.

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