Analytic geometry proof with triangle.

[ghostfirefox]

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 $$\vec{BC}$$.

 

[skeeter]

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?

 

[HallsofIvy]

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 $$\vec{BD}$$?

 

[ghostfirefox]

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?