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

thanks

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- Apr 26th 2006, 08:35 PMjacstriangle area proof
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

thanks

- Apr 26th 2006, 08:49 PMearbothQuote:

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.

Greetings

EB

to b) - Apr 26th 2006, 08:58 PMjacs
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

jacs