Proving using vectors, that if a median is also a height, then the triangle is isosceles.

Question

*Better wording would be very helpful.

Thanks in advance for any help.

score 1 · Answer 1 · answered Nov 03 '13 at 16:46

consider the triangle with one vertex at the origin, and the other two vertex being the points representing the vectors $a$ and $b$. the median is perpendicular to the base. This can be written in terms of $a$ and $b$ to be $$(b-a)\cdot\frac{a+b}{2}=0$$ because dot product of two parpendicular vectors is $0$. Taking modulus of both sides we can show $|b|=|a|$.

score 0 · Accepted Answer · answered Nov 03 '13 at 16:52

0

Let the vertices be the origin $O$, and the points $A$ and $B$. Let $u$ be the vector $OA$, and let $v$ be the vector $OB$. Then if $M$ is the midpoint of $AB$, the vector $OM$ is $\frac{u+v}{2}$.

Because the median is an altitude, we have $\left(\frac{u+v}{2}\right)\cdot (u-v)=0$. Thus $(u+v)\cdot(u-v)=0$.

Expand. We get $u\cdot u-v\cdot v=0$, so $u\cdot u=v\cdot v$.

But $u\cdot u$ is the square of the length of $OA$, and $v\cdot v$ is the square of the length of $OB$. It follows that the length of $OA$ is equal to the length of $OB$, and therefore the triangle is isosceles.

answered Nov 03 '13 at 16:52

André Nicolas

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Thanks you. Only why is it the OM=${u+v}\over 2$? – richard Nov 03 '13 at 17:09
Recall that if you have two points $(a,b)$ and $(c,d)$ in the plane, their midpoint is $(\frac{a+c}{2}. \frac{b+d}{2})$. If you want to do it in vector form, calculate $\frac{u+v}{2}-u$ and $v-\frac{u+v}{2}$. After simplification, they are equal, which shows $\frac{u+v}{2}$ is the midpoint. – André Nicolas Nov 03 '13 at 17:19

score 0 · Answer 3 · answered Nov 03 '13 at 16:59

Let $A,B,C$ be the vertices and $M$ be the centre of $AB$. Then $CM$ is the median and height of the triangle. Then by Pythagorian Theorem which stands for vector spaces with inner product we have that $|AC|^2=|AM|^2+|CM|^2$ and $|CB|^2=|BM|^2+|CM|^2=|AM|^2+|CM|^2$ and thus $|AC|=|CB|$. Suppose what is inside $||$ is a vector.

Alternative answer from what our friends wrote

score 0 · Answer 4 · answered Nov 03 '13 at 17:03

Let $\;\Delta ABC\;$ be the triangle, and let $\;M\;$ be the midpoint of $\;BC\;$ . Denote vectors $\;u:=\vec{AB}\;,\;v:=\vec{AC}\;$. Since

$$\vec{BC}=v-u\implies \vec{BM}=\vec{MC}=\frac12(v-u)$$

and we're given

$$\vec{AM}\perp\vec{BC}\;,\;\;\text{and we also know that}\;\;\vec{AM}=\frac12(u+v)$$

But then

$$\vec{AM}\perp\vec{BC}\iff 0=\left\langle\;\frac12(u+v),v-u\;\right\rangle=\frac12\left(||v||^2-||u||^2\right)\iff$$

$$\iff ||v||=||u||\iff \text{the triangle's isosceles}$$

Proving using vectors, that if a median is also a height, then the triangle is isosceles.

4 Answers4