Proving Diagonal Trisection in a Parallelogram using Vector Calculus

[link2001]

the problem goes:

ABCD is a parallelogram in which points P and Q are the midpoints of sides BC and CD, respectively. Use vector calculus to prove that AP and AQ trisect the diagonal BD at the points E and F.


...A _________B
.../...F.../
.../...E.../P
D/________/C
...Q
(Imagine lines from D->B, A->Q and A->P, where AQ passes through E and AP passes through F)


I set the Problem Up like this:
DE = x(DA + AB)
EF = y(DA + AB)
FB = z(DA + AB)

After this I get stuck running around in circles trying to make subsitutions to show that x, y & z = 1/3.

Ideas anyone?

 

[member 553083]

We can use the fact that P and Q are the midpoints of sides BC and CD to solve this problem. Start by noting that BD = DA + AB, since these are the two sides of a parallelogram. Then, we can say that DE = x(BD), EF = y(BD), and FB = z(BD). Now, since P is the midpoint of BC, we can say that BP = CP. By the same logic, AQ = AD. This means that DE + EF = BP and EF + FB = AQ. Combining our equations, we get x(BD) + y(BD) = BP and y(BD) + z(BD) = AQ. Since BP = CP and AQ = AD, we can substitute these in for the right hand side of the equations. We then have: x(BD) + y(BD) = CP, and y(BD) + z(BD) = AD. Now, we can add the two equations together to get (x + y + z)(BD) = CP + AD. Since BD = DA + AB, this becomes (x + y + z)(DA + AB) = CP + AD. Finally, we can divide both sides of the equation by (DA + AB), which gives us x + y + z = (CP + AD)/(DA + AB). Since CP = AD, we can simplify this to x + y + z = 1/2. Since x, y, and z must all be positive numbers, the only way for their sum to equal 1/2 is if they are each equal to 1/3. Thus, DE trisects BD at E, and EF trisects BD at F.