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Question
If D and E are the midpoints of the sides AB and AC of a triangle ABC, prove that `vec"BE" + vec"DC" = 3/2vec"BC"`
Solution
Let O be the origin.
Let `vec"a", vec"b", vec"c"` the position vectors of the points A, B and C respectively
Then `vec"OA" = vec"a", vec"OB" = vec"b", vec"OC" = vec"c"`
Given D is the midpoint of AB
∴ `vec"OD" = (vec"OA" + vec"OB")/2`
`vec"OD" = (vec"a" + vec"b")/2`
Also given E is the midpoint of AC
∴ `vec"OE" = (vec"OA" + vec"OC")/2`
`vec"OE" = (vec"a" + vec"c")/2`
`vec"BE" = vec"OE" - vec"OB"`
= `(vec"a" + vec"c")/2 - vec"b"`
`vec"BE" = (vec"a" + vec"c" - 2vec"b")/2`
`vec"DC" = vec"OC" - vec"OD"`
= `vec"c" - (vec"a" + vec"b")/2`
`vec"DC" = (2vec"c" + vec"a" - vec"b")/2`
`vec"BE" + vec"DC" = (vec"a" + vec"c" - 2vec"b")/2 + (2vec"c" - vec"a" - vec"b")/2`
= `(vec"a" + vec"c" - 2vec"b" + 2vec"c" - vec"a" - vec"b")/2`
= `(3vec"c" - 3vec"b")/2`
= `3/2(vec"c" - vec"b")`
= `3/2(vec"OC" - vec"OB")`
`vec"BE" + vec"DC" = 3/2vec"BC"`
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