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Question
Let A, B, and C be the vertices of a triangle. Let D, E, and F be the midpoints of the sides BC, CA, and AB respectively. Show that `vec"AD" + vec"BE" + vec"CF" = vec0`
Solution
Let ABC be the triangle.
D, E, F are the midpoints of the sides BC, CA, AB respectively.
Let `vec"a", vec"b"` and `vec"c"` be the position vectors of the vertices A, B, C respectively.
Then `"OA" = vec"a"`
`vec"OB" = vec"b"`
`vec"OC" = vec"c"`
D is the midpoint of BC
∴ Position vector of D is = `(vec"OB" + vec"OC")/2`
`vec"OD" = (vec"b" + vec"c")/2`
E is the midpoint of CA
∴ Position vector of E is = `(vec"OC" + vec"OA")/2`
`vec"OE" = (vec"c" + vec"a")/2`
F is the midpoint of AB
∴ Position vector of F is = `(vec"OA" + vec"OB")/2`
`vec"OF" = (vec"a" + vec"b")/2`
`vec"AB" = vec"OD" - vec"OA"`
= `(vec"b" +vec"c")/2 - vec"a"`
`vec"AD" = (vec"b" + vec"c" - 2vec"a")/2` ......(1)
`vec"BE" = vec"OE" - vec"OB"`
= `(vec"c" +vec"a")/2 - vec"b"`
`vec"BE" = (vec"c" + vec"a" - 2vec"b")/2` .......(2)
`vec"CF" = vec"OF" - vec"OC"`
= `(vec"a" + vec"b")/2 - vec"c"`
`vec"CF" = (vec"a" + vec"b" - 2vec"c")/2` .......(3)
From equation (1), (2) and (3) we have
`vec"AD" +vec"BE" + vec"CF" = (vec"b" + vec"c" - 2vec"a")/2 + (vec"c" + vec"a" - 2vec"b")/2 + (vec"a" + vec"b" - 2vec"c")/2`
= `(vec"b" + vec"c" -2vec"a" + vec"c" + vec"a" - 2vec"b" + vec"a" + vec"b" - 2vec"c")/2`
= `(2vec"a" + 2vec"b" + 2vec"c" - 2vec"a" - 2vec"b" - 2vec"c")/2`
`vec"AD" + vec"BE" + vec"CF" = 0/2` = 0
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