Please help to find the angle

Hello, OnlyMath!

I did a lot of trig and came with a surprisingly simple answer.
I suspect that there is an elegant geometric solution . . . somewhere.

Quote:

$ABC$ is a triangle where the point $D$ is on $BC$
. . such that: . $\angle BAD = 60^o,\;\;\angle B=20^o,\;\' AB = CD$

Find $\angle C.$

Code:

B o * * * 20 * * * D a * 100 o * * 80 * a * 60 * x * * * @ * A o * * * * * * * o C

We have: . $\angle BAD = 60^o,\;\angle B = 20^o \quad\Rightarrow\quad \angle BDA = 100^o,\;\angle ADC = 80^o$

Let $a = AB = CD,\;x = AD,\;\theta = \angle C \quad\Rightarrow\quad \angle DAC = 100-\theta$

$\text{In }\Delta ABD\!:\;\dfrac{x}{\sin20^o} \:=\:\dfrac{a}{\sin100^o} \quad\Rightarrow\quad \dfrac{x}{a} \:=\:\dfrac{\sin20^o}{\sin100^o}$ .[1]

$\text{In }\Delta ADC\!:\;\dfrac{x}{\sin\theta} \:=\:\dfrac{a}{\sin(100^o-\theta)} \quad\Rightarrow\quad \dfrac{x}{a} \:=\:\dfrac{\sin\theta}{\sin(100^o-\theta)}$ .[2]

Equate [1] and [2]:

. . $\dfrac{\sin20}{\sin100} \:=\:\dfrac{\sin\theta}{\sin(100-\theta)} \quad\Rightarrow\quad \sin20\sin(100-\theta) \;=\;\sin100\sin\theta<br />$

. . $\sin20\bigg(\sin100\cos\theta - \cos100\sin\theta\bigg) \;=\;\sin100\sin\theta$

. . $\sin20\sin100\cos\theta - \sin20\cos100\sin\theta \;=\;\sin100\sin\theta$

. . $\sin100\sin\theta + \sin20\cos100\sin\theta \;=\;\sin20\sin100\cos\theta$

. . $\bigg(\sin100 + \sin20\cos100\bigg)\sin\theta \;=\;\sin20\sin100\cos\theta$

. . . . . . . $\dfrac{\sin\theta}{\cos\theta} \;=\;\dfrac{\sin20\sin100}{\sin100 + \sin20\cos100}$

Therefore: . $\tan\theta \:=\:0.363970234 \quad\Rightarrow\quad \theta \:=\:20^o$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Fascinating!

$\Delta ABC$ is isosceles: . $AB \,=\,AC\,=\,a$

Also, $\Delta ADC$ is isosceles: . $AC\,=\,CD\,=\,a$

Is there a more direct way to derive those facts?