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 PQ/BD.
P.S. ASDF is wrong
Triangles BMP and DCP are similar, so BP/DP = BM/CD. Since quadrilateral ABCD is a parallelogram, and M is the midpoint of AB, we have BM = AB/2 = CD/2, so BP/DP = BM/CD = 1/2. But BP + DP = BD, so BP = BD/3. Next, triangles DNQ and BCQ are similar, so DQ/BQ = DN/BC. Since quadrilateral ABCD is a parallelogram, and N is the midpoint of AD, we have DN = AD/2 = BC/2, so DQ/BQ = DN/BC = 1/2. But BQ + DQ = BD, so DQ = BD/3.
Finally, PQ/BD} = (BD - BP - DQ) / BD = 1 - BP/BD - DQ/BD = 1 - 1/3 - 1/3 = 1/3.
Your answer is 1/3.
1/3 is correct
To find the ratio PQ/BD, we need to determine the lengths of PQ and BD.
Let's note some important properties of parallelograms:
1. Opposite sides of a parallelogram are congruent.
This means that AB = CD and AD = BC.
2. The diagonals of a parallelogram bisect each other.
This means that M is also the midpoint of CD, and N is also the midpoint of BC.
3. The diagonals of a parallelogram divide it into four congruent triangles.
This means that △BPC is congruent to △DQC, and △BQC is congruent to △DPC.
Now, we can find the lengths of PQ and BD.
Since M is the midpoint of AB, we know that AM = MB. Similarly, since N is the midpoint of AD, we know that DN = NC.
From property 1, we have AB = CD and AD = BC. Hence, AM + MD = CD, and DN + NC = BC.
Since M is the midpoint of CD and N is the midpoint of BC, we can express CD and BC in terms of AM and DN:
CD = 2 * AM and BC = 2 * DN
Now, let's consider the triangles △BPC and △DPC.
△BPC and △DPC share the same base PC and have the same height BD, which is the length of the diagonal BD.
Therefore, the area of △BPC is equal to the area of △DPC.
Using the formula for the area of a triangle (0.5 * base * height), we have:
0.5 * PC * BD = 0.5 * PC * BD
Since the bases and heights are equal, the ratio of the sides opposite the shared base (PQ/BD) must also be equal.
So, we conclude that PQ/BD = 1.
Therefore, PQ is equal in length to BD, and the ratio PQ/BD is 1.
This is an AoPS question...
stop cheating!!!
Solution 1: Triangles ADE and BFE are similar since \overline{AD}\parallel\overline{BF}. Furthermore, the ratio of their areas is 9:1, so AE/BE = 3. Then CD = AB = AE + BE = 3BE + BE = 4BE. Triangles FEB and FDC are similar, so the area of triangle FDC is (DC/BE)^2 \cdot [FEB] = 16 \cdot 1 = 16.
Therefore, the area of quadrilateral BEDC is [FDC] - [FEB] = 16 - 1 = 15, so the area of parallelogram ABCD is [ABCD] = [ADE] + [BEDC] = 9 + 15 =
Ans=24