# I am finding a contradiction in my vector proof. What am I doing wrong?

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?

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 $$MC$$.

• I see. Looks like my understanding of vectors needs some refinement – user71207 Jan 20 at 12:03

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.

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.

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 $$\vec{AM}=\vec{M}-\vec{0}=\frac{\mathbf{a}+\mathbf{b}}{2}$$