ABCD is a trapezium, AB and CD are parallel sides, M is the
midpoint of AD. Prove that the area of the triangle BMC is half
the area of the trapezium. I let A be the origin, then the
vectors of AB, AC, and AD respectively are b, c, d. So vector
AM= Area of the trapezium= vector CM= vector CB= My problem is I
am stuck here, I think it is just some rearrangement of the
vectors, but I am no good at dot and cross-products.