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Question
If D, E, F are the midpoints of the sides BC, CA, AB of a triangle ABC, prove that `bar(AD) + bar(BE) + bar(CF) = bar0`.
Theorem
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Solution
Let `bara, barb, barc, bard, bare, barf` be the position vectors of the points A, B, C, D, E, F respectively.
Since D, E, F are the midpoints of BC, CA, AB respectively, by the midpoint formula
`bard = (barb + barc)/2, bare = (barc + bara)/2, barf = (bara + barb)/2`
∴ `bar(AD) + bar(BE) + bar(CF) = (bard - bara) + (bare - barb) + (barf - barc)`
`= ((barb + barc)/2 - bara) + ((barc + bara)/2 - barb) + ((bara + barb)/2 - barc)`
`= 1/2barb + 1/2barc - bara + 1/2barc + 1/2bara - barb + 1/2bara + 1/2barb - barc`
`= (bara + barb + barc) - (bara + barb + barc) = bar0`.
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