2. Find the length of segment HE. Points E, D, and H are the midpoints of the sides of triangle TUV. UV = 80, TV = 100, and HD = 80
First, let's draw a diagram to help us visualize the problem:
[asy]
pair T, U, V, H, E, D;
T = (0,0);
U = (100,0);
V = (40,50);
H = (70,0);
D = (70,25);
E = (20,25);
draw(T--U--V--cycle);
draw(T--D);
draw(T--E);
draw(U--H);
label("$T$", T, SW);
label("$U$", U, SE);
label("$V$", V, N);
label("$H$", H, S);
label("$D$", D, NE);
label("$E$", E, NW);
label("80", T--V, NW);
label("100", U--V, NE);
label("80", T--U, S);
label("80", T--D, S);
label("80", HD, S);
[/asy]
We can see that segment HD is parallel to and half the length of segment UV, and segment HE is perpendicular to and half the length of segment TV. Therefore, triangle UTH is similar to triangle VHE.
Let's set up a ratio with the corresponding side lengths: $\frac{HT}{TV} = \frac{HU}{VE}$. Substituting in the known lengths, $\frac{HT}{100} = \frac{70}{VE}$.
We can solve this equation to find that $VE = \frac{70}{100} \cdot HT = \frac{7}{10} \cdot HT$.
Since we know that segment HD is half the length of segment UV, $HD = \frac{1}{2} \cdot 80 = 40$. Since segment HE is half the length of segment TV, $VE = \frac{1}{2} \cdot 100 = 50$. Therefore, $HT = HV + VT = HD + VE = 40 + 50 = 90$.
Finally, substituting back into the equation we found earlier, $VE = \frac{7}{10} \cdot HT = \frac{7}{10} \cdot 90 = \boxed{63}$.