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.

Well, well, well, we have a parallelogram puzzle on our hands, don't we? Let's try to solve it with a touch of humor!

Now, let's start by acknowledging something - M is the midpoint of AB and N is the midpoint of AD? Sounds like M and N are really good friends with the midpoints, doesn't it? They should totally start a club!

Moving on, we have BD intersecting CM and CN at P and Q, respectively. Wow, this parallelogram is becoming quite the intersection party, isn't it? Who knew geometry could be so social?

Now, we're asked to find PC/BD. Ah, division, my dear friend! It's like the mathematical version of splitting the bill at a restaurant. So, let's get calculating!

Since M is the midpoint of AB, we know that MC is equal to MB. Similarly, since N is the midpoint of AD, we have NC equals ND. So far, so good!

Now, let's take a look at triangle CPD. We notice that triangles CMB and CND are similar. Why? Well, because they share an angle at C and CM/MB = CN/ND due to being midpoints. It's like they're mirror images of each other, just a little bit smaller!

Now, if triangles CMB and CND are similar, then the corresponding sides are proportional. This means that PC/BD = CM/MB = CN/ND.

But wait, since MB = MC and ND = NC, we can substitute these equalities in. Ah, the joy of substitution!

So, PC/BD = CM/MC = CN/NC.

And guess what? CM/MC = 1 and CN/NC = 1 because they're equal! Isn't that delightful?

Therefore, PC/BD = 1/1 = 1.

So, my friend, the ratio of PC to BD is 1. It's as simple as that! I hope my quirky sense of humor helped make this parallelogram problem a little more enjoyable for you.

To find the ratio PC/BD, we can use the property that in a triangle, if a line is drawn parallel to one side, it divides the other two sides proportionally.

Step 1: Draw the parallelogram ABCD with diagonal BD intersecting CM and CN at P and Q respectively.

Step 2: Draw line MP parallel to CD.

Step 3: Since MP is parallel to CD, it divides the sides BC and AD proportionally. Therefore, we can write:

PM/MB = PC/BD

Step 4: We know that M is the midpoint of AB, so PM = MB.

Step 5: Substituting PM = MB into the equation from step 3, we have:

MB/MB = PC/BD

Step 6: Simplifying, we get:

1 = PC/BD

So, the ratio PC/BD is 1.

To find PC/BD, we can use the properties of similar triangles.

First, let's label the given points and segments on the diagram for convenience:

Let ABCD be a parallelogram with M as the midpoint of AB and N as the midpoint of AD. Let CM intersect diagonal BD at point P, and CN intersect diagonal BD at point Q. We need to find the ratio PC/BD.

Now, let's consider the two similar triangles formed within the parallelogram.

Triangle CPB is similar to triangle QAB.
Triangle CQN is similar to triangle QDC.

Since triangle CPB is similar to triangle QAB, we can write the following proportion of corresponding sides:

PC / QA = BC / AB

Similarly, since triangle CQN is similar to triangle QDC, we can write:

QN / CD = CQ / BC

We know that QN = 1/2 AD, CD = AD, and AB = BD (since diagonals of a parallelogram bisect each other).

Substituting these values into the above proportions, we can simplify them as follows:

PC / (1/2 AD) = BC / BD
CQ / AD = BC / BD

To eliminate the fractions, we can multiply both equations by AD and BD, respectively:

PC * BD = (1/2 AD) * BC
CQ * AD = BC * BD

From the second equation, we can solve for CQ:

CQ = (BC * BD) / AD

Notice that BC and BD are common factors in both equations. We can eliminate them by dividing the second equation by the first equation:

(CQ * AD) / (PC * BD) = ((BC * BD) / AD) / ((1/2 AD) * BC)
(CQ * AD) / (PC * BD) = 2

Therefore, the ratio PC/BD is equal to 2 or PC : BD = 2 : 1.

So, PC is twice as long as BD in this parallelogram.

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 $\frac{PQ}{FH}=\boxed{\frac13}$.