Point $D$ In $\triangle{ABC}$ such that $\angle{DBC}=\angle{DCB}=10^{\circ}, \angle{BAD}=20^{\circ}, \angle{DAC}=40^{\circ}$, show that $AD=AC$. Wonder if there is pure geometric approach to prove this statement? Thanks
For trigonometric solution, apply Ceva's theorem in angles:
Let $\angle{ABD}=x$
$$
\begin{multline}\nonumber
\shoveleft \dfrac{\sin20}{\sin40} \dfrac{\sin(100-x)}{\sin10} \dfrac{\sin10}{\sin x}=1 \\
\shoveleft \implies \dfrac{\sin(100-x)}{2\sin x \cdot \cos20}=1 \\
\shoveleft \implies \cos(x-10)=2\sin x \cdot \cos20\\
\shoveleft \implies \cos x \cdot \cos10+\sin x \cdot \sin10=2\cos20 \cdot \sin x\\
\shoveleft \implies \tan x = \dfrac{\cos10}{2\cos20-\sin10}=\dfrac{\sin30 \cdot \cos10}{\cos20 -\sin30 \cdot \sin10}\\
\shoveleft =\dfrac{\sin30 \cdot \cos10}{\cos20-\tfrac{1}{2}(\cos20 + \cos40)}=\dfrac{\sin30 \cdot \cos10}{\tfrac{1}{2}(\cos20-\cos40)}\\
\shoveleft = \dfrac{\sin30 \cdot \cos10}{\cos30 \cdot \cos10}=\tan30 \implies x=30^{\circ} \implies \angle{ACD}=70^{\circ} \implies AC=AD \blacksquare
\end{multline}
$$
Best Answer
Let $E$ be a point on the circumcircle of $ABD$ such that $DB=DE$. Then $\angle DBE = \angle BAD = 20^\circ$. Note that $B, C, E$ lie on a circle with center $D$. Hence $\angle CDE = 2\angle CBE = 2\cdot 30^\circ = 60^\circ$. It follows that triangle $CDE$ is equilateral. In particular, $DE=CE$. Since $\angle DAE = 20^\circ$, we also have $\angle EAC = 20^\circ$. This shows that $E$ lies both on the internal bisector of $\angle DAC$ and on the perpendicular bisector of $CD$. This can only happen if $AD=AC$ or if $AD\neq AC$ and $E$ is the midpoint of the arc $CD$ of the circumcircle of $ADC$ (but this is not the case since $E$ and $A$ lie on the same side of the line $CD$). Therefore $AD=AC$.