Let ABCD be a parallelogram. Let M be the midpoint of AB and N be the midpoint of AD. Diagonal BD intersects CM and CN at P and Q respectively find PC/BD.

Answer 1

Solution 1:
[asy]
unitsize(1 cm);

pair E, F, G, H, M, N, P, Q;

E = (1,2);
F = (5,2);
G = (4,0);
H = (0,0);
M = (E + F)/2;
N = (E + H)/2;
P = extension(F,H,G,M);
Q = extension(F,H,G,N);

draw(E--F--G--H--cycle);
draw(F--H);
draw(G--M);
draw(G--N);

label("$E$", E, NW);
label("$F$", F, dir(0));
label("$G$", G, SE);
label("$H$", H, SW);
label("$M$", M, dir(90));
label("$N$", N, W);
label("$P$", P, NNE);
label("$Q$", Q, dir(90));
[/asy]

Triangles $FMP$ and $HGP$ are similar, so
\[\frac{FP}{HP} = \frac{FM}{GH}.\]Since quadrilateral $EFGH$ is a parallelogram, and $M$ is the midpoint of $EF$, we have $FM = EF/2 = GH/2$, so
\[\frac{FP}{HP} = \frac{FM}{GH} = \frac{1}{2}.\]But $FP + HP = FH$, so $FP = FH/3$.

Next, triangles $HNQ$ and $FGQ$ are similar, so
\[\frac{HQ}{FQ} = \frac{HN}{FG}.\]Since quadrilateral $EFGH$ is a parallelogram, and $N$ is the midpoint of $EH$, we have $HN = EH/2 = FG/2$, so
\[\frac{HQ}{FQ} = \frac{HN}{FG} = \frac{1}{2}.\]But $FQ + HQ = FH$, so $HQ = FH/3$.

Finally,
\[\frac{PQ}{FH} = \frac{FH - FP - HQ}{FH} = 1 - \frac{FP}{FH} - \frac{HQ}{FH} = 1 - \frac{1}{3} - \frac{1}{3} = \boxed{\frac{1}{3}}.\]
Solution 2:
[asy]
unitsize(1 cm);

pair E, F, G, H, M, N, P, Q, X;

E = (1,2);
F = (5,2);
G = (4,0);
H = (0,0);
M = (E + F)/2;
N = (E + H)/2;
P = extension(F,H,G,M);
Q = extension(F,H,G,N);
X = extension(F,H,G,E);

draw(E--F--G--H--cycle);
draw(F--H);
draw(G--M);
draw(G--N);
draw(E--G);

label("$E$", E, NW);
label("$F$", F, dir(0));
label("$G$", G, SE);
label("$H$", H, SW);
label("$M$", M, dir(90));
label("$N$", N, W);
label("$P$", P, NNE);
label("$Q$", Q, dir(90));
label("$X$", X, dir(80));
[/asy]

Draw in diagonal $\overline{EG}$, meeting $\overline{FH}$ at point $X$. Since the diagonals of a parallelogram bisect each other, $EX = XG$. Therefore, $\overline{FX}$ and $\overline{GM}$ are medians of $\triangle EFG$, so $P$ is the centroid of $\triangle EFG$. Therefore,
\[\frac{PX}{FX} = \frac13.\]Similarly, $Q$ is the centroid of $\triangle EGH$, so
\[\frac{QX}{HX} = \frac13.\]Combining these results,
\[PQ = PX + QX = \frac{FX}3 + \frac{HX}3 = \frac{FH}3,\]so $\frac{PQ}{FH}=\boxed{\frac13}$.