CEMC Pascal, 2025

Since $AB=AD$, then $\triangle ABD$ is isosceles and $\angle ADB=\angle ABD=80\degree$.

Since $\angle BDC$ is a straight
angle, then $\angle ADC=180\degree-\angle ADB=180\degree-80\degree=100\degree$.

$\triangle ADC$ is also isosceles
(since $AD=CD$), and so $\angle CAD=\angle ACD$.

The sum of the angles in $\triangle ADC$ is $180\degree$, and so
$\angle ADC+\angle CAD+\angle ACD=180\degree$ or $100\degree+2\times\angle ACD=180\degree$
or $2\times\angle ACD=80\degree$, and
so the measure of $\angle ACD$ is
$40\degree$.

$\triangle ABD$ is isosceles
with $AB=AD$, and so $\angle ADB=\angle ABD=80\degree$. The
measure of $\angle BDC$ is 180$\degree$ since it is a straight angle.
Thus, $\angle ADC=180\degree-\angle ADB=180\degree-80\degree=100\degree$.

$\triangle ADC$ is isosceles with
$AD=DC$, and so $\angle ACD=\angle CAD$. The sum of the
three angles in $\triangle ADC$ is
$180\degree$, and so $\angle ACD+\angle CAD=180\degree-\angle ADC=180\degree-100\degree=80\degree$. Since $\angle ACD+\angle CAD=80\degree$ and $\angle ACD=\angle CAD$, then $\angle ACD=\dfrac{80\degree}{2}=40\degree$.