solution to the first one:

Let D, E, F to be the mid-points of BC,CA and AB respectively. given that the medians AD,BE,CF are all equal.

The medians are concurrent at G.

now GD=GE=GF=(1/3)AB.

this shows that G is the circumcentre of triangle DEF.

also EF is parallel to BC. so by basic proportionality theorem AG bisects the side EF. Since G is the circumcentre, the line joining the mid-point of EF to G is perpendicular to EF. since EF is parallel to BC the line AG extented(which id the median AD) meets BC at right angles. So the median AD is perpendicular to BC. similarly the other two medians also meet the corresponding sides at right angles. this will easily lead to the conclusion that the triangle ABC is equilateral.