Question

Show by vector method that the perpendicular bisectors of sides of a triangle are concurrent

Answer 1

Let A, B, C be the vertices of a triangle with position vectors a, b, c . Let D, E, F be the midpoints of BC, CA, AB respectively. Let S be the point of intersection of the perpendiculars drawn at D, E to BC, CA respectively.

let vector{OS = r}

now vector{OD = (b+c)/2, OE = (c+a)/2,OF = (a+b)/2}

vector{SD ⊥ BC, SE ⊥ CA}

vector{SD ⊥ BC => SD.BC = 0 => (OD - OS).(OC - OB)} = 0

=> vector[{(b+c)/2 - r}. (c-d)] = 0

=> vector{(b+c).(c-b)/2 = r.(c-b)}

=> vector{(c²-b²)/2 = r.c-r.b} (1)

Similarly vector{SE ⊥ CA => (a²-c²)/2 r.a - r.c} (2)

(1) + (2) => vector{(a²-b²)/2 = r.a - r.b} => vector{(a+b).(a-b)/2 = r.(a-b)}

=> vector{(a+b)/2.(a-b) - r.(a-b)} = 0 

=> vector[{(a+b)/2.(a-b)}] = 0

=> vector{(OF - OS).(OA - OB)} = 0 

=> vector{SF.BA} = 0

=> vector{SF ⊥ BA}

Perpendicular of F to AB passes through S.

Perpendicular bisectors of sides are concurrent.