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 $A D = C D = B D$ . Now it is obvious that $C D = B D = \frac{1}{2} B C$ . Since D is the midpoint of BC.
Consider
$\begin{matrix} - - \to A D = - - \to A B + - - \to B D = - - \to A B + \frac{1}{2} - - \to B C = - - \to A B + \frac{1}{2} (- - \to B A + - - \to A C) - - \to A B - \frac{1}{2} - - \to A B + \frac{1}{2} - - \to A C = \frac{1}{2} (- - \to A B + - - \to A C) ∴ {(- - \to A D)}^{2} = \frac{1}{4} {(- - \to A B + - - \to A C)}^{2} = \frac{1}{4} ({- - \to A B}^{2} + 2 - - \to A B \cdot - - \to A C + {- - \to A C}^{2}) A D^{2} = \frac{1}{4} (A B^{2} + 0 + A C^{2}) = \frac{1}{4} B C^{2} (b y p y t h a g o r a s t h e o r e m) ∴ A D = \frac{1}{2} B C S o w e h a v e A D = B D = C D \end{matrix}$

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Prove that in a right-angled triangle, the mid-point of the hypotenuse is equidistant from the vertices.

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