Vectors: Prove that if median in a triangle is perpendicular to the corresponding base, then the triangle is isosceles. Prove using vector methods that if a median in a triangle is perpendicular to the corresponding base, then the triangle is isosceles.
I know the basic idea. The dot product of the median vector and that of the base vector is $0$. Using that we have to prove the lengths of the sides containing the median are equal, but I don't know how to go about doing that.
Hints please.
Edit
My working: (as Blue suggested)
The equation for$\vec c$ is $ \vec r = (1-t)\vec a + t \vec b $ and the equation for $\vec m$ is $\vec r = \lambda\Big(\dfrac {\vec a + \vec b} 2\Big)$
Setting the dot product to zero gives this:
$$ \lambda/ 2 (1 + \vec a \cdot \vec b) = 0$$
What can I conclude from this?
 A: As a hint, think about where the median line intersects the base, using that information (and that which you already stated) you should be able to reach the conclusion.
A: Edit. Re-written to review fundamental facts for OP's benefit.


Facts
  
  
*
  
*The vector from point $P$ to point $Q$ is given by $\vec{PQ} = Q-P$.
  
*The midpoint of $PQ$ is $\frac{1}{2}\left(P+Q\right)$
  

(To derive the second fact, note that one gets to the midpoint by starting at $P$ and traveling along the half-vector towards $Q$. This puts the midpoint at
$$P + \frac{1}{2}\vec{PQ} = P + \frac{1}{2}\left(Q-P\right) = \frac{1}{2}\left( 2 P + Q - P \right) = \frac{1}{2}\left( P + Q \right)$$
as claimed.)

Now, let the $\triangle AOB$ have "base" $AB$, with $M$ the midpoint of that base. Situate vertex $O$ at the origin. From the above, we see that sides of the triangle, and the median in question, determine these vectors:
$$\vec{OA} = A - O = A \qquad \vec{OB} = B - O = B \qquad \vec{AB} = B - A$$
$$\vec{OM} = M-O = M = \frac{1}{2}\left( A + B \right)$$
We're investigating what happens when the median and the base are perpendicular, which corresponds to the condition:
$$\vec{OM}\cdot\vec{AB} = 0$$
Substituting our formulas for the vectors, this becomes
$$\frac{1}{2} \left( A + B \right)\cdot \left( B - A \right) = 0$$
Clearing the multiplied $\frac{1}{2}$ from both sides, we can expand the dot product and simplify thusly:
$$\begin{align}
0 &= A\cdot B - A\cdot A + B \cdot B - B\cdot A \\
&= |OB|^2 - |OA|^2
\end{align}$$ 
Consequently,
$$|OA|^2 = |OB|^2$$
and the triangle is isosceles.
A: You should be able to prove that the two triangles into which the median cuts the original triangle have one common side (the median), and they have another equal side - the base. The angle between the median and the base is equal in the two triangles.
Therefore we have equality of two sides and the included angle, which is sufficient to show that the triangles are congruent and the other corresponding sides and angles are equal.
To use vector methods, choose origin the midpoint of the base. Call the vector to the apex $\vec a$ and the vector to the right-hand base vertex $\vec b$. We then have $\vec a\cdot \vec b=0$.
The length of the right-hand side is $(\vec a - \vec b)\cdot(\vec a - \vec b)=\vec a \cdot \vec a   - 2\vec a\cdot \vec b +\vec b \cdot \vec b=|a|^2+|b|^2$
The the vector to the left-hand base vertex is $-\vec b$ so the length of the left-hand side is similarly $(\vec a + \vec b)\cdot(\vec a + \vec b)$
