Suppose Eric is E years old and Dong is D years old.

Let D-E = d

(d is the difference between Eric and Dong's age. It shall remain the same no matter how old they both get.)

Take up the first statement:

When Dong's age was half the sum of their present ages, i.e (D+E)/2,

Eric's age was (D+E)/2 - d

When Dong would be twice as old as that he would be:

2[(D+E)/2 - d] years old

When Dong is that many years old, Eric will be 2[(D+E)/2 - d] - d years old

And that is equal to Dong's present age. Therefore:

D = 2 [ (D+E)/2 - d] - d

Substituting E = D-d

D = 2[ (D+D-d)/2 - d] - d

Simplifying

D = 4d .......(1)

Now the second statement

In 10 years, Eric will be E+10 years old

Half that age is (E+10)/2

When Eric was (E+10)/2 years old Dong was (E+10)/2 + d years old

And that is equal to Eric's present age

E = (E+10)/2+d

Substituting E = D-d

D-d = (D-d+10)/2+d

Simplifying,

D =3d+10 ..............(2)

Solving (1) and (2),

4d = 3d+10

d=10

And D = 4d = 40, Dong's age - The answer.

Nice problem. I set up a system of equations that had nine variables and nine equations. Don't freak out! It was still a very simple system to solve. There were four possible dates that were discussed. I denoted their ages during these dates by $D_1,D_2,D_3,D_4,E_1, E_2, E_3, E_4$ . I then let $x$ be the difference in their ages; that is, $D_i=x+E_i$ (this is four equations). Next, I followed each segment one at a time to set up the remaining equations. "Dong is as old as Eric will be": $D_1=E_2$ . "when Dong is twice as old as Eric was": $D_2=2E_3$ . "when Dong's age was half the sum of their present ages": $D_3=\frac{1}{2}(D_1+E_1)$ . "Eric is as old as Dong was": $E_1=D_4$ . "when Eric was half the age he will be 10 years from now": $E_4=\frac{1}{2}(E_1+10)$ . Now eliminate all the D's by using the first four equations (and then forget about those first four). After doing that, the equations become: $x+E_1=E_2,x+E_2=2E_3,x+E_3=\frac{1}{2}(x+E_1+E_1),E_1=x+E_4,E_4=\frac{1}{2}(E_1+10)$ . Remember that we need to find $x$ and $E_1$ so that we can determine $D_1$ . Substitute the first equation into the second. Then the equations of interest become: $x+(x+E_1)=2E_3, x+E_3=\frac{1}{2}x+E_1,E_1=x+E_4,E_4=\frac{1}{2}E_1+5$ . Then substitute the fifth equation into the fourth: $2x+E_1=2E_3,E_3=-\frac{1}{2}x+E_1, E_1=x+\frac{1}{2}E_1+5$ . Then substitute the second equation into the first: $2x+E_1=2(-\frac{1}{2}x+E_1), E_1=x+\frac{1}{2}E_1+5$ . Simplifying: $E_1=3x, E_1=2x+10$ . So, $3x=2x+10\Rightarrow x=10$ , and $E_1=3x=30$ . And, so, $D_1=E_1+x=30+10=40$ .

I'll follow the first valid/understandable solution writer.