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Prove the midpoint of the hypotenuse of a right-angled triangle is equidistance from the three vertices by vector methods.

I have drawn out a triangle with the right angle at A, then going clockwise, vertices B, and C.

By definition of perpendicular vectors, $(\vec{AM} + \vec{MC})\cdot(\vec{AM}-\vec{BM})=0$. Then use the fact that $\vec{BM} = \vec{MC}$ to conclude the proof.

However, lets try to define $\vec{AC}$ = a and $\vec{AB}$ = b

$\vec{BC}$ = a $-$ b so $\vec{BM} = \vec{MC} = \frac{1}{2}($a$-$b)

But then for $\vec{AM} =$ a $- \vec{MC} = \frac{1}{2} ($a$+$b) which is obviously not equal to BM or MC. Why is that happening?

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There is no contradiction: by equidistant we mean that $|\vec {AM}| = |\vec {BM}| = |\vec {CM}|$.

Indeed we have $\vec {AM} = \frac12(\mathbf a + \mathbf b)$. Thus:

$$|\vec {AM}| = \frac12\sqrt{(\mathbf a + \mathbf b)\cdot (\mathbf a + \mathbf b)}=\frac12\sqrt{\mathbf a \cdot \mathbf a+\mathbf a \cdot \mathbf b+2\mathbf a \cdot \mathbf b} = \frac12\sqrt{|\mathbf a|^2+|\mathbf b|^2} = \frac12|BC|$$

so the length of $AM$ is equal to $BM$ and $MC$.

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    $\begingroup$ I see. Looks like my understanding of vectors needs some refinement $\endgroup$ – user71207 Jan 20 at 12:03
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What you are asked to prove is that $\vec{AM}$, $\vec{BM}$, and $\vec{CM}$ have the same norm (or length if you prefer), not that they are equal as vectors.

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So, you have shown:

$\vec{BM} = \vec{MC} = \frac{1}{2}(\mathbf{a}-\mathbf{b})$

and

$\vec{AM} = \frac{1}{2}(\mathbf{a}+\mathbf{b})$

All that remains is to show AM = BM (not that the vectors are equal, simply their magnitude). Here, we can use the observation that $\mathbf{a} \cdot \mathbf{b} = 0$, and the result that $|\mathbf{v}|^2=\mathbf{v}\cdot\mathbf{v}$

$BM^2 = (\frac12)^2(\mathbf{a}-\mathbf{b})\cdot(\mathbf{a}-\mathbf{b})=(\frac12)^2(a^2 -2\mathbf{a}\cdot\mathbf{b}+b^2)=(\frac12)^2(a^2 +b^2)$

$AM^2 = (\frac12)^2(\mathbf{a}+\mathbf{b})\cdot(\mathbf{a}+\mathbf{b})=(\frac12)^2(a^2 +2\mathbf{a}\cdot\mathbf{b}+b^2)=(\frac12)^2(a^2 +b^2)$

So AM = BM.

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Note that $\vec{BM}$ and $\vec{MC}$ point in same direction (along the hypotenuse). While $\vec{AM}$ points in a different direction (from right vertex to midpoint of hypotenuse). So $\vec{AM}$ must be different from $\vec{BM}$ or $\vec{MC}$.

But when you calculate their lengths, you'll obtain $$|\vec{AM}|=|\vec{BM}|=|\vec{MC}|$$

Things would have been easier had you also defined position vector of vertex $A$ as origin. Then position vector of $M$ could easily be seen as $$\vec{M}=\frac{\mathbf{a}+\mathbf{b}}{2}$$ that is $$\vec{AM}=\vec{M}-\vec{0}=\frac{\mathbf{a}+\mathbf{b}}{2}$$

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