Nice problem. $AB=c,AC=b,BC=a$ First part. The altitude of $B$ is smaller or equal to the median $B$ so the median in $C$ is smaller or equal than the altitude in $C$. it can't be smaller so it must be equal therefore the equality is for both altitudes and medians in $B$ and $C$ so $AB=BC$ and $BC=CA$ therefore all angles must be 60. Second part first let $M$ be midpoint of $AB$ and $H$ be the feet of $A$ on $BC$ and $K$ the midpoint of $BH$ now $CMK$ is right with $CM=2*MK$ so $\angle MCA=30$ similarly if $N$ is the midpoint of $AC$ $=\angle ABN=30$ let $T$ be symmetric to $B$ wrt $N$ triangles $BCG$ and $BCT$ are similar(where $G$ is the centroid of $ABC$) since we can express the medians by lenghts of the triangle the privous similarity will give us that $2c^2=a^2+b^2$ now using cosine law on $\triangle BNA$(it has an angle of 30) and using the privous side equality to eliminate $c$ we compute that either $a^2-7b^2=0$ or $a^2-b^2=0$ second case makes $CA=CB$ and with $\angle BCM=30$ it means $\angle ACB=60$ triangle is equilateral($\angle ABC=60$) first case gives $c=2a$ and by cosine law gives $\angle ABC=120$. we can construct a triangle that satisfiesthe given conditions first by constructing the right triangle $BCG$($\angle BCG=30$) than extending $CG$ to $M$ where $\angle MBG=30$ and than the perp from $M$ to $BG$ gives $N$ now $A$ is the intersection of $BM$ and $CN$. the only thing u need to prove is that $MN$ is the midline of $ABC$ which is true since $MN$=$BM/2=BC/2$ and $MN||BC$ so $60$ and $120$ are all possible values for $\angle ABC$