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Question

Using vectors, prove that the mid-point of the hypotenuse of a right-angled triangle is equidistant from its vertices.

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Solution

Let ABC be a right triangle, right angled at A, Let D be the midpoint of the hypotenuse BC. We show that AD=CD=BD. Now it is obvious that CD=BD=12BC. Since D is the midpoint of BC.
Consider
AD=AB+BD=AB+12BC=AB+12(BA+AC)AB12AB+12AC=12(AB+AC)(AD)2=14(AB+AC)2=14(AB2+2ABAC+AC2)AD2=14(AB2+0+AC2)=14BC2(bypythagorastheorem)AD=12BCSowehaveAD=BD=CD

1117561_1129696_ans_2f57cce8135541d7b0f0d80bdea966fb.png

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