THE 68th ROMANIAN MATHEMATICAL OLYMPIAD

a) If $M$ is on the hypotenuse of the triangle $ABC$, with a right angle in $A$, with $m(\angle B)=30^{\circ }$ such that $m(\angle BAM)=30^{\circ }$, then the triangle $MAC$ is equilateral. Hence $MA=CM=CA=\frac{1}{2}BC$, and $MB=\frac{1}{2}BC$, also.

b) We construct $D^{\prime }\in (AD)$, such that $CA=C{D}^{\prime }$, $D^{\prime }\in \text{Int}(ABC)$, $m(\angle AC{D}^{\prime })=40^{\circ }$, $m(\angle D^{\prime }CB)=20^{\circ }$. If $CG\perp A{D}^{\prime }$, $G\in (A{D}^{\prime })$, and $D^{\prime }F\perp BC$, $F\in (BC)$, then $△C{D}^{\prime }G\equiv △C{D}^{\prime }F$ (H.A.), so $D^{\prime }G=D^{\prime }F$.

$D^{\prime }E\perp AB$, $E\in (BC)$, $G$ is the midpoint of $[A{D}^{\prime }]$, $m(\angle D^{\prime }AE)=30^{\circ }$, $D^{\prime }F=D^{\prime }G=D^{\prime }E$, hence $D^{\prime }$ is on the angle bisector of the angle $B$, $m(\angle D^{\prime }BA)=10^{\circ }$. $D^{\prime }\in (AD)$, $D^{\prime }\in (BD)$, $D^{\prime }=D$, hence $m(\angle ACD)=40^{\circ }$.