The Excercise goes as follows:
I have $2$ - $4$ sided figures, $ABCD$ and $ADCM$.
$M$ is the centre of the circumscribed circle of $ABCD$ with $DB$ being its diameter.
additoinally, $ABD$ is defined as $42 ^\circ$ and $EC$ has the same length as $EM$.
Calculate the angle of $CAD$
I was able to calculate most of the lower angles but I just can't figure out how to calculate any of the upper ones. I know I have to use the identical lenght of EM and EC, but the only ifnormation that gives me, is that EMC is a same sided triangle.
Can anybody makes sonse of all this? I just kind of seams like I have no idea how to get one of the uppre angles using one of the lower ones.
Since $EC=EM,$ triangle $\triangle CEM$ is isoceles, and $\triangle AMC$ is isoceles because two of the side are radii of the circle.
From $\triangle CEM$ we get $\angle ECM=\angle EMC.$ From $\triangle AMC$ we get $\angle MAC=\angle MCA.$ Also observe that because $E$ is on the line segment $\overline{AC},$ we have $\angle ECM=\angle MCA.$
So we have several angles that are all the same size. So that we don't get confused by all the names referring to different locations in the figure, let the size be $\alpha.$ That is, we have $$\angle ECM=\angle EMC=\angle MCA=\angle MAC=\alpha.$$
Now consider the triangle $\triangle AMC$ again. One of its angles is $\angle MAC=\alpha,$ one is $\angle MCA=\alpha,$ and the third is $\angle AMC = \angle AME+\angle EMC=\angle AME+\alpha.$ The three angles of an triangle must have sum $180,$ so $$ \alpha + \alpha + (\angle AME+\alpha) = 180. $$ You can find $\angle AME$ as an exterior angle of the isoceles triangle $\triangle AMB$ or as the measure of the arc of the circle intercepted by the inscribed angle $\angle ABD.$ Then solve for $\alpha.$
Finally, use the fact that $\angle CAD+\angle CAM+\angle MAB=\angle BAD=90,$ so knowing the measures of $\angle CAM$ and $\angle MAB,$ find $\angle CAD.$