Question 1123348
Draw this one.  Let the equal legs be x, so median is 0.5x (30-60-90 triangle) and the third leg of the bisected triangle is 0.5x sqrt (3)
Perimeter of the two triangles is 2(3x+x sqrt(3))
perimeter of the large triangle is 4x+2xsqrt (3)
therefore, 6x+2x sqrt(3)-20=4x+2x sqrt(3)
2x=20
x=10 cm
the sides of ABC are 20-20-10 sqrt (3) or 40+10 sqrt(3) or about 57.32
each of the two triangles has perimeter 20-10-5 sqrt(3) or about 38.66.  Two of them have perimeter 77.32
The last leg of ABC is 10 sqrt(3) cm or 17.32 cm.