prove that AZ=2ZC
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To tackle this problem I decided to try using a graph. Point A of the triangle is at the origin (0,0); point B is at (2X,0) and point X at (X,0), so that X is the midpoint of AB. The third point of the triangle at C is (a,b), an arbitrary point. We know that Y is halfway along CX so that means Y is ((X+a)/2,b/2), because its x coord along base AB is halfway between X (X,0) and C (a,b) and its y coord is halfway between B and C.

We can write an equation for AC: y=bx/a. And we can write an equation for BZ because it's a continuation of BY, and we know the coords of B and Y. The slope of the line is (b/2)/((X+a)/2-2X)=b/(a-3X), the difference of the y values divided by the difference of the x values of the two points. The equation for the line becomes y=bx/(a-3X)+k, where k is the y intercept we have to find. Since point B (2X,0) is on the line we can substitute the coords to find k: 0=2Xb/(a-3X)+k, so k=-2Xb/(a-3X) and the equation of the line BZ is y=bx/(a-3X)-2Xb/(a-3X)=(b/(a-3X))(x-2X).

The two lines AC and BZ intersect at bx/a=(b/(a-3X))(x-2X). We can divide through by b to get: x/a=(x-2X)(a-3X). Multiply through by a(a-3X): x(a-3X)=a(x-2X), so -3xX=-2aX, and x=2a/3. The y value  can be found by substitution: y=bx/a=2b/3 because the a's cancel out. The point C is (a,b). Therefore the point Z is (2a/3,2b/3) and this implies that AZ lies 2/3 the way along AC and AZ=2ZC. QED.


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