MHB Analytic geometry proof with triangle.

ghostfirefox
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Point D divides side AC, of triangle ABC, so that |AD|: |DC| = 1:2. Prove that vectors \vec{BD} = 2/3 \vec{BA} + 1/3 \vec{BC}.
 
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ghostfirefox said:
Point D divides side AC, of triangle ABC, so that |AD|: |DC| = 1:2. Prove that vectors \vec{BD} = 2/3 \vec{BA} + 1/3 \vec{BC}.

Have you made a sketch?
 
You titled this "Analytic geometry proof ..." so I would set up a coordinate system so that the origin is at the vertex "B" of the triangle and the x-axis lies along on side BC. Then B= (0, 0), A= (x, y), and C= (z, 0) for some numbers x, y, and z. Since D lies on AC such that "|AD|:|DC|= 1:2", D= ((2x+ z)/3, 2y/3). Now, what are the vectors \vec{BA}, \vec{BC}. and \vec{BD}?
 
HallsofIvy said:
You titled this "Analytic geometry proof ..." so I would set up a coordinate system so that the origin is at the vertex "B" of the triangle and the x-axis lies along on side BC. Then B= (0, 0), A= (x, y), and C= (z, 0) for some numbers x, y, and z. Since D lies on AC such that "|AD|:|DC|= 1:2", D= ((2x+ z)/3, 2y/3). Now, what are the vectors \vec{BA}, \vec{BC}. and \vec{BD}?

I have a question how you calculated the x coordinate of point D?
 
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