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

Open in App

Solution

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 = 12 BC. Since D is the midpoint of BC.
consider, $- - \to A D = - - \to A B + - - \to B D = - - \to A B + \frac{1}{2} - - \to B C = - - \to A B + \frac{1}{2} (- - \to B A + - - \to A C)$
$- - \to A D = - - \to A B - \frac{1}{2} - - \to A B + \frac{1}{2} - - \to A C = \frac{1}{2} (- - \to A B + - - \to A C)$
$∴$ ${(- - \to A D)}^{2} = \frac{1}{4} {(- - \to A B + - - \to A C)}^{2}$
${(- - \to A D)}^{2} = \frac{1}{4} {(- - \to A B + - - \to A C)}^{2} = \frac{1}{4} ({- - \to A B}^{2} + {- - \to A C}^{2} + 2 - - \to A B . - - \to A C)$
i.e. , ${A D}^{2} = \frac{1}{4} ({A B}^{2} + {A C}^{2} + 0) s i n c e (- - \to A B ⊥ - - \to A C)$
= ${A D}^{2} = \frac{1}{4} {B C}^{2}$
$A D = \frac{1}{2} B C$
So we have AD = BD = CD
Hence proved.

Suggest Corrections

Similar questions

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

Join BYJU'S Learning Program