perpendicular bisectors concurrent?

Question

i'm having problems proving (using the dot product) that the perpendicular bisectors of the sides of a triangle are concurrent. this is what i've tried:

assume true, then concurrent at a point D. therefore, letting "." representing the dot product,

(D-.5(A+B) . (A-B) = 0
(D-.5(B+C) . (B-C) = 0
(D-.5(C+A) . (C-A) = 0.

i don't really know what to do next.... any suggestions? or other approaches maybe? thanks.

Answer 1

I've never seen the proof with dot porducts, and I don't see it immediately, but I will say that you are on the wrong track. When proving that something is true, if you assume that it is true, and then arrive at some other true result, you've proven nothing. If you arrive at a falsehood, then you do disprove the assumption, but otherwise, nothing can be accomplished.

Therefore, perhaps it is best to assume that two perendicular bisectors intersect at a point distinct from the intersection of another pair, and then try to reach a contradiction.

Answer 2

Another approach would be to draw two perpedicular bisectors(l and n) that intersect at point T. The two lines will intersect because they are perpendicular to two lines that intersect. let l be the perpendicular bisector of BC and n be the perpendicular bisector of BA of the triangle ABC.

Now the point T is equidistant from points B and C and the Point T is also equidistant from the points A and B.
Thus the point T is equidistant from poits A and C.

Now draw the line m as the perpendicular pisector of AC.
the point T must lie on m because m is the locus of all points equidistant from A and C and T is one of those points.

Therefore the perpendicular bisectors are concurrent at the point T. T is the circumcenter of the triangle and is the center of the circle that passes through all three vertices of the triangle, thus circumscribing it.