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Prove that the mid-point of the hypoten...

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

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Here, `angleCAB = 90^@,` let D be the mid-point of hypotenuse, we have
BD = DC
AB = AD+DB
AC = AD+DC = AD+BD
Since, `angleBAC =90^@ AB bot AC `

(AD + DB). (AD +BD) = 0
(AD - BD). (AD+BD)=0
`:. AD^2 - BD^2 = 0 `
AD = BD also BD = DC
`:'` D is mid point of BC
Thus, |AD| = |BD| = |DC|. Hence, the result.
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