Question

prove by vector method that the midpoint of a right angled triangle is equidistant from its vertices

Answer 1

Answer :-

Let ABC be a right triangle, righte angled at A. Let D be the midpoint of the hypotenuse BC. We have to show that AD = CD = BD. Now it is obvious that CD = BD = 1212 BC. Since D is the midpoint of BC.

Consider AD−→−=AB−→−+BD−→−=AB−→−+12BC−→−=AB−→−+12(BA−→−+AC−→−)AD→=AB→+BD→=AB→+12BC→=AB→+12(BA→+AC→)

=AB−→−−12AB−→−+12AC−→−=12(AB−→−+AC−→−)=AB→−12AB→+12AC→=12(AB→+AC→)

∴(AD−→−)2=14(AB−→−+AC−→−)2=14(AB−→−2+2AB−→−.AC−→−+AC−→−2)∴(AD→)2=14(AB→+AC→)2=14(AB→2+2AB→.AC→+AC→2)

i.e., AD2=14[AB2+0+AC2][sinceAB⊥AC]AD2=14[AB2+0+AC2][sinceAB⊥AC]

=14BC2=14BC2 ( by pythagoras theorem)

∴AD=12BC∴AD=12BC

So we have AD = BD = CD