[tex]d(M,DC)=5\sqrt3 \text{ evident }\\
\text{ Folosind teorema celor 3 perpendiculare rezulta: }\\
MD\perp\AD,MC\perp BC\Rightarrow d(M,AD)=MD,d(M,BC)=MC\\
MC=MD \text{ si se afla cu teorema lui Pitagora; }\\
MC=MD=\sqrt{5^2+(5\sqrt3)^2}=10\\
\text{ Considerand $F$ mijlocul lui $AB$ si folosind aceeasi teorema rezulta}\\
MF\perp AB \Rightarrow d(M,AM)=MF\\
MF^2=\sqrt{(5\sqrt3)^2+10^2}=5\sqrt7\\
[/tex]
[tex]d(M,AC)=\text{lungimea inaltimii }\Delta MAC\\
\text{Folosind teorema lui Pitogora obtinem:}\\
AC=10\sqrt2 , MC=10, AE=5\sqrt5,MA=10\sqrt2 [/tex]
[tex]\text{In }\Delta MAC: h_M\cdot AC=h_A\cdot MC\Rightarrow h_M=\frac{h_A\cdot MC}{AC}
\\
\text{Dar $h_A$ se poate afla tot cu Pitagora:}
\\ h_A=\sqrt{(10\sqrt2)^2-5^2}=5\sqrt7\\
h_M=\frac{5\sqrt7\cdot10 }{10\sqrt2}=\frac{5\sqrt14}{2}[/tex]
