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Show that the perpendicular bisectors of the sides of a triangle are concurrent.

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Let D,E,F be the midpoints of the sides BC,CA and AB of `triangleABC`.
Let the perpendicular bisectors of the sides BC and AC meet each other in the points O. Choose O as the origin and let `bar(a),bar(b),bar(c),bar(d),bar(e),bar(f)` be the position vectors of the points A,B,C,D,E,F respectively.
Here we have to prove that `bar(OF)=bar(f)` is perpendicular to `bar(AB)=bar(b)-bar(a)`.
By the midpoint formula, `bar(d)=(bar(b)+bar(c))/(2),bar(e)(bar(c)+bar(a))/(2),bar(f)=(bar(a)+bar(b))/(2)`
Now, `bar(OD)=bar(d)` is perpendicular to `bar(BC)=bar(c)-bar(b)`.
`:.bar(d)*(bar(c)-bar(b))=0" ":.((bar(b)+bar(c))/(2))*(bar(c)-bar(b))=0`
`:.(bar(c)+bar(b))*(bar(c)-bar(b))=0`
`:.bar(c)*bar(c)+bar(b)*bar(c)-bar(c)*bar(b)-bar(b)*bar(b)=0`
`:.c^(2)-b^(2)=0" ":.c^(2)=b^(2). . . [becausebar(c)*bar(c)=c^(2),bar(b)*bar(b)=b^(2)andbar(c)*bar(b)=bar(b)*bar(c)]` . . .(1)
Also, `bar(OE)=bar(e)` is perpendicular to `bar(AC)=bar(c)-bar(a)`
`:.bar(e)*(bar(c)-bar(a))=0`
`:.((bar(c)+bar(a))/(2))*(bar(c)-bar(a))=0" ":." as above "c^(2)=a^(2)` . . . .(2)
From (1) and (2), we get,
`b^(2)=a^(2)" ":.b^(2)-a^(2)=0`
`:.(bar(b)+bar(a))*(bar(b)-bar(a))=0`
`:.((bar(b)+bar(a))/(2))*(bar(b)-bar(a))=0`
`:.bar(f)=bar(OF)` is perpendicular to `bar(b)-bar(a)=bar(AB)`.
`:.` the perpendicular bisectors of the sides of `DeltaABC` are concurrent.
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