Ask a math problem about vectors and triangles? O is always a point in the triangle ABC, and there are: OA ^ 2 + BC ^ 2 = ob ^ 2 + Ca ^ 2 = OC ^ 2 + AB ^ 2, proving: ab ⊥ OC. Note: in the condition, OA, BC, ob, CA, OC, AB are the form of module length of vector; in proving, AB and OC are the form of vector

Ask a math problem about vectors and triangles? O is always a point in the triangle ABC, and there are: OA ^ 2 + BC ^ 2 = ob ^ 2 + Ca ^ 2 = OC ^ 2 + AB ^ 2, proving: ab ⊥ OC. Note: in the condition, OA, BC, ob, CA, OC, AB are the form of module length of vector; in proving, AB and OC are the form of vector

OA ^ 2 + BC ^ 2 = ob ^ 2 + Ca ^ 2 = OC ^ 2 + AB ^ 2oa ^ 2-ob ^ 2 = Ca ^ 2-bc ^ 2 (OA + OB) (oa-ob) = (Ca + CB) (ca-cb) BA (OA + OB) - BA (Ca + CB) = 0ba (OA + ob + AC + BC) = 0ba (OC + OC) = 0ab · 2oC = 0ab · OC = 0