Trapezium vector problem
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.
Yes, I am supposed to use vectors.
Refer the attached image in#2. Draw MN parallel to AB.
Let the vector CB be b.
Now the area of the triangle MNC =
Area of the triangle BMC = 2*MNC =
= 1/2*[ares(ABC) + ares(DBC)]
But Area(ABC) = area(ADC)
So Area(BMC) = 1/2*[area(ADC) + ares(DBC)] = 1/2[areaABCD]
N is the midpoint of BC.
Originally Posted by arze