In Δ ABC, P is a point on the edge of BC, and | BP | = 2 | PC |, and D is the midpoint of AC. AP and BD intersect at point O. we use vectors AB and AC to represent vector Ao Use the solution of the three points collinear; a = λ B + (1 - λ) C

In Δ ABC, P is a point on the edge of BC, and | BP | = 2 | PC |, and D is the midpoint of AC. AP and BD intersect at point O. we use vectors AB and AC to represent vector Ao Use the solution of the three points collinear; a = λ B + (1 - λ) C

In Δ ABC, P is a point on the edge of BC, and | BP | = 2 | PC |,
∴AP=(1/3)AB+(2/3)AC,
And D is the midpoint of AC,
Let the vector Ao = λ AB + (1 - λ) ad = λ AB + (1 - λ) / 2 * AC,
AP and BD intersect at point o,
∥ AP ∥ Ao, AB, AC are not collinear,
∴λ/（1/3）=[(1-λ)/2]/(2/3),
∴4λ=1-λ,λ=1/5,
The vector Ao = (1 / 5) AB + (2 / 5) AC

In △ ABC, P is a point on the edge of BC, and BP = 2 pc. we use vector AB and vector AC to represent vector AP

Vector AP = vector AB + vector BP vector BP = 2|3 vector BC vector BC = vector AC - vector ab
Vector BP = 2|3 vector AC vector AB vector AP = 1|3 vector AB + 2|3 vector AC