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Prove by vector method, The mid point of the hypotenuse of a right angled tringle is equidistant from its vertices.

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  • If $ \overrightarrow a \perp \overrightarrow b$ then $ \overrightarrow a.\overrightarrow b=0$ and for nonzero vectors if $ \overrightarrow a.\overrightarrow b=0 \Rightarrow \overrightarrow a \perp \overrightarrow b.$
  • For any two vectors $ \hat a \: and \: \hat b$ $(\hat a + \hat b)^2=(\hat a)^2+2\hat a.\hat b+(\hat b)^2=a^2+2\hat a.\hat b+b^2$ $(\hat a-\hat b)^2=a^2-2\hat a.\hat b+b^2$ $(\hat a+\hat b).(\hat a-\hat b)=a^2-b^2$
  • By $ \Delta$ law of vectors if $ \overrightarrow a+\overrightarrow b=\overrightarrow c \: or \: \overrightarrow a+\overrightarrow b=-\overrightarrow c$ then the vectors form the sides of a $ \Delta$
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 = $ \large\frac{1}{2}$ BC. Since D is the midpoint of BC.
Consider $ \overrightarrow {AD}=\overrightarrow {AB}+\overrightarrow {BD}=\overrightarrow {AB}+ \large\frac{1}{2}\overrightarrow {BC}=\overrightarrow {AB}+ \large\frac{1}{2}(\overrightarrow {BA}+\overrightarrow {AC})$
$ = \overrightarrow {AB}-\large\frac{1}{2}\overrightarrow {AB}+\large\frac{1}{2}\overrightarrow {AC}=\large\frac{1}{2}(\overrightarrow {AB}+\overrightarrow {AC})$
$ \therefore (\overrightarrow {AD})^2 = \large\frac{1}{4} (\overrightarrow {AB}+\overrightarrow {AC})^2=\large\frac{1}{4}(\overrightarrow {AB}^2+2\overrightarrow {AB}.\overrightarrow {AC}+\overrightarrow {AC}^2)$
i.e., $ AD^2=\large\frac{1}{4} [ AB^2+0+AC^2]\: [ since\: AB \perp AC]$
$ = \large\frac{1}{4} BC^2 $ ( by pythagoras theorem)
$ \therefore AD = \large\frac{1}{2}BC$
So we have AD = BD = CD
answered Jun 4, 2013 by thanvigandhi_1
edited Jun 6, 2013 by meena.p

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