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.

Answer 1

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

Answer 2

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.

Answer 3

stop cheating!!!

Answer 4

P.S. ASDF is wrong

Answer 5

This is an AoPS question...

Answer 6

1/3 is correct