SAT Question 6d99b141: Lines, Angles, and Triangles answer and explanation with Desmos solution

It's given that in the figure, $AC=CD$. Thus, triangle $ACD$ is an isosceles triangle and the measure of angle $CDA$ is equal to the measure of angle $CAD$. The sum of the measures of the interior angles of a triangle is $180^{\circ }$. Thus, the sum of the measures of the interior angles of triangle $ACD$ is $180^{\circ }$. It's given that the measure of angle $ACD$ is $104^{\circ }$. It follows that the sum of the measures of angles $CDA$ and $CAD$ is ${\left(180-104\right)}^{\circ }$, or $76^{\circ }$. Since the measure of angle $CDA$ is equal to the measure of angle $CAD$, the measure of angle $CDA$ is half of $76^{\circ }$, or $38^{\circ }$. The sum of the measures of the interior angles of triangle $BDE$ is $180^{\circ }$. It's given that the measure of angle $EBC$ is $45^{\circ }$. Since the measure of angle $BDE$, which is the same angle as angle $CDA$, is $38^{\circ }$, it follows that the measure of angle $DEB$ is ${\left(180-45-38\right)}^{\circ }$, or $97^{\circ }$. Since angle $DEB$ and angle $AEB$ form a straight line, the sum of the measures of these angles is $180^{\circ }$. It's given in the figure that the measure of angle $AEB$ is $x^{\circ }$. It follows that $97+x=180$. Subtracting $97$ from both sides of this equation yields $x=83$.