$ABCD$ is a convex quadrilateral with $\angle CAB=60^\circ$, $\angle CAD=20^\circ$, $\angle ABD=50^\circ$, $\angle DBC=10^\circ$. Find $\angle ACD$. Can you help me solve this Olympiad angles problem?

Let $ABCD$ be a convex quadrilateral such that: $\widehat{CAB}=60^\circ$; $\widehat{CAD}=20^\circ$; $\widehat{ABD}=50^\circ$; $\widehat{DBC}=10^\circ$.
Calculate $\widehat{ACD}$.

Thank you so much for your help

Let $E$ be the intersection point of lines $AC$ and $BD$.
Then:
$\angle BEC=60^\circ+50^\circ=110^\circ$
$\begin{align} \angle BCE&= 180^\circ-110^\circ - 10^\circ \\ &=60^\circ \\ &= \angle{BAC} \end{align}$
So that means $\Delta ABC$ is equilateral and hence:
$AB=AC \tag 1$
You can also deduce that:
$\begin{align} \angle BDA &= 180^\circ - 60^\circ - 20^\circ - 50^\circ \\ &= 50^\circ \\ &= \angle ABD \end{align}$
So $\Delta BAD$ is isosceles and hence:
$AB=AD \tag 2$
Combining $(1)$ and $(2)$ leads to:
$AD=AC$
Meaning that:

 $\triangle DAC$ is isosceles

I believe you can finish up now.
