Answer
$\sqrt{17},\ 2\sqrt{5},\ \sqrt{29}$
Work Step by Step
Midpoint of A and B:
$M_{AB}=\displaystyle \left(\frac{0+6}{2},\frac{0+0}{2}\right)=(3,0)$
$d(C,M_{AB})=\sqrt{(0-4)^{2}+(3-4)^{2}}=\sqrt{(-4)^{2}+(-1)^{2}}=\sqrt{16+1}=\sqrt{17}$
$M_{AC}=\displaystyle \left(\frac{0+4}{2},\frac{0+4}{2}\right)=(2,2)$
$d(B,M_{AC})=\sqrt{(2-6)^{2}+(2-0)^{2}}=\sqrt{(-4)^{2}+2^{2}}=\sqrt{16+4}=\sqrt{20}=2\sqrt{5}$
$M_{BC}=\displaystyle \left(\frac{6+4}{2},\frac{0+4}{2}\right)=(5,2)$
$d(A,M_{BC})=\sqrt{(2-0)^{2}+(5-0)^{2}}=\sqrt{2^{2}+5^{2}}=\sqrt{4+25}=\sqrt{29}$