# Proof pythagoras theorem with dot product + distance

I want to proof $d(A.B)^2=d(A,C)^2+d(B,C)^2$ for

with $(\vec a-\vec c) \bullet (\vec b - \vec c)=0$. I applied the definitions of distance and got

$d(A,B)^2=d(A,C)^2+d(B,C)^2 \Leftrightarrow ||\vec b - \vec a||^2= || \vec c - \vec b||^2+||\vec c - \vec a||^2$ but now I don't know how to proceed.

If someone could help me out, it would be great.

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Note that $\|c-b\|^2=(c-b)\bullet (c-b)$ and $\|c-a\|^2=(c-a)\bullet (c-a)$. Since $(a-c)\bullet(b-c)=0$, you can rewrite the right hand side your equation as:$$\|c-b\|^2+\|c-a\|^2=\|c-b\|^2+2(a-c)\bullet (b-c)+\|c-a\|^2$$Try simplifying this using the properties of the dot product.

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perhaps something like $\|c-b\|^2+\|c-a\|^2=\|c-b\|^2+2(a-c)\bullet (b-c)+\|c-a\|^2 \Leftrightarrow \|c\|^2+\|b\|^2-2\|c\| \bullet \|b\|+\|c\|^2+\|a\|^2-2\|c\|\bullet \| a\| +2(a-c) \bullet (b-c)$ ? – ulead86 Apr 24 '13 at 10:05

$$\vec{b}-\vec{a}=(\vec{b}-\vec{c})+(\vec{c}-\vec{a})$$

$$\implies (\vec{b}-\vec{a})^{2}=\left[(\vec{b}-\vec{c})+(\vec{c}-\vec{a})\right]^{2}=(\vec{b}-\vec{c})^{2}+(\vec{c}-\vec{a})^{2}+2(\vec{b}-\vec{c})\cdot(\vec{c}-\vec{a})$$

It is known that $(\vec{a})^{2}=||a||^{2}$ and $(\vec{b}-\vec{c})\perp(\vec{c}-\vec{a})$ or$(\vec{b}-\vec{c})\cdot(\vec{c}-\vec{a})=0$

Therefore,

$$||\vec b - \vec a||^2= || \vec b - \vec c||^2+||\vec c - \vec a||^2$$ or,$$d(A,B)^2=d(A,C)^2+d(B,C)^2$$

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