A given parallelogram ABCD, M is the intersection point of the diagonals of AC and BD, for any given point O, Proof that .

To prove that OM is perpendicular to AB, we need to show that the dot product of OM and AB is zero.

Let's first find the coordinates of M. We know that M is the intersection point of AC and BD, so we can find M by solving the system of equations:

x/3 + y/2 = 1 (equation of AC)
x/2 + y/3 = 1 (equation of BD)

Solving these equations, we get:

x = 6/5
y = 6/5

Therefore, the coordinates of M are (6/5, 6/5).

Now, let's find the vectors OM and AB.

OM = M - O = (6/5 - x, 6/5 - y) - (x, y) = (1/5 - x, 1/5 - y)

AB = B - A = (3, 0) - (0, 2) = (3, -2)

The dot product of OM and AB is:

OM · AB = (1/5 - x)(3) + (1/5 - y)(-2)

Substituting the values of x and y, we get:

OM · AB = (1/5 - 6/5)(3) + (1/5 - 6/5)(-2) = 0

Therefore, OM is perpendicular to AB.