Question

prove that mid point of the hypotenuse of a right angle triangle is equidistant from its vertices A(0,0),B(2a,0),C(02b)​

Answer 1

Answer: :

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 result.

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