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Any Point Equidistant from the End Points of a Segment Lies on the Perpendicular Bisector of the Segment

If Δ ABC is...

Question

If $Δ A B C$ is right angled at $B$ , then $A B^{2} + B C^{2} = A C^{2}$ . Prove using only vectors.

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Solution

$\to A B + \to B C = \to A C$
$\to A B . \to B C = 0$ as AB is perpendicular to BC
$| \to A C |^{2} \Rightarrow | \to A B + \to B C |^{2} = | \to A B + \to B C | . | \to A B + \to B C | \Rightarrow | \to A B |^{2} + | \to B C |^{2} + 2 | \to A B | | \to B C |$
$| \to A C |^{2} = | \to A B |^{2} + | \to B C |^{2}$

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