In triangle ABC, X is the midpoint of AB and Y is the midpont of AC.
XC is joined. Prove
a) ∆AXY = ∆XYC in area
b) ∆AXC = half ∆ABC in area
c) Find what fraction the area of ∆AXY is of the ∆ABC
Hello,Originally Posted by jacs
to a). Both triangles have the same base: XY and the same height, because XY divides the height of ABC in 2 equal parts.
to b) XC divides (ABC) in 2 equal areas.
to c) Because area(AXY) = area(XYC) and area(ABC) = 2 * area(AXC) therefore area(ABC) = 4 * area(AXY)
(The three midparallels of a triangle divide the triangle in 4 congruent triangles, similar to the great one)
I've attached a copy of your diagram to demonstrate how I calculated.
thanks for that EB, i see it now and i should have seen it before (duh to me) I actually cheated and used trig area of triangle rule for part a, utilising
sin a = sin (180 - a) which i knew was not the way to go but couldnt see the other way.
thanks for that