### Instruction:
Let $ABCD$ be a parallelogram. We have that $M$ is the midpoint of $AB$ and $N$ is the midpoint of $BC.$ The segments $DM$ and $DN$ intersect $AC$ at $P$ and $Q$, respectively. If $AC = 15,$ what is $QA$? [asy]
pair A, B, C, D, pM, pN, O, P, Q;
A = (25, 0) * dir(-20);
B = (15, 0) * dir(60);
C = (25, 0) * dir(160);
D = (15, 0) * dir(-120);
pM = 0.5 * A + 0.5 * B;
pN = 0.5 * B + 0.5 * C;
O = 0.25 * A + 0.25 * B + 0.25 * C + 0.25 * D;
P = 0.33 * C + 0.67 * A;
Q = 0.67 * C + 0.33 * A;
draw(A--B--C--D--cycle);
draw(A--C);
draw(pM--D);
draw(pN--D);
label("$A$", A, S);
label("$B$", B, NE);
label("$C$", C, NW);
label("$D$", D, S);
label("$M$", pM, NE);
label("$N$", pN, NW);
label("$P$", P, N);
label("$Q$", Q, NE);
[/asy]

### Response:

Solution 1: The segment $BD$ begs to be drawn, so we start there: [asy]
pair A, B, C, D, pM, pN, O, P, Q;
A = (25, 0) * dir(-20);
B = (15, 0) * dir(60);
C = (25, 0) * dir(160);
D = (15, 0) * dir(-120);
pM = 0.5 * A + 0.5 * B;
pN = 0.5 * B + 0.5 * C;
O = 0.25 * A + 0.25 * B + 0.25 * C + 0.25 * D;
P = 0.33 * C + 0.67 * A;
Q = 0.67 * C + 0.33 * A;
draw(A--B--C--D--cycle);
draw(A--C);
draw(B--D);
draw(pM--D);
draw(pN--D);
label("$A$", A, S);
label("$B$", B, NE);
label("$C$", C, NW);
label("$D$", D, S);
label("$M$", pM, NE);
label("$N$", pN, NW);
label("$P$", P, N);
label("$Q$", Q, NE);
[/asy] We can clearly see that now we have triangles $ABD$ and $CBD,$ and $MD,$ $ND,$ and $AC$ are medians to one or more of the triangles. That means that $P$ and $Q$ are the centroids of triangles $ABD$ and $CBD,$ respectively. Since $AC = 15,$ that means $CQ = 5,$ since the median from $C$ to $BD$ is half the length of $AC,$ or $7.5,$ and $CQ$ must be $\frac{2}{3}$ of that, or $5.$ Therefore, $QA = AC - CQ = 15 - 5 = 10.$

Solution 2: Since $ABCD$ is a parallelogram, $\overline{AD}$ and $\overline{BC}$ are parallel with $\overline{AC}$ and $\overline{DN}$ as transversals.  So $\angle DAQ = \angle NCQ$ and $\angle ADQ = \angle CNQ$, and so $\triangle ADQ$ and $\triangle CNQ$ are similar by AA similarity.

Also, we know opposite sides of a parallelogram are congruent, so $AD = BC$.  Since $N$ is a midpoint of $\overline{BC}$, we have $CN = \dfrac{AD}2$.  By similar triangles, \[ \dfrac{AQ}{CQ} = \dfrac{AD}{CN} = 2, \]so $AQ = 2CQ.$  Since $AQ + CQ = AC = 15$, we have $CQ = 5$ and $AQ = 10.$