Prove that the mid-point of hypotenuse of right triangle is equidistant from its vertices
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Here, ∠CAB=90∘, let D be the mid-point of hypotenuse, we have
BD = DC
AB = AD+DB
AC = AD+DC = AD+BD
Since, ∠BAC=90∘AB⊥AC
(AD + DB). (AD +BD) = 0
(AD - BD). (AD+BD)=0
∴AD2−BD2=0
AD = BD also BD = DC
∵ D is mid point of BC
Thus, |AD| = |BD| = |DC|.
Hence, the mid-point of hypotenuse of right triangle is equidistant from its vertices.
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