Prove, by vector methods, that in an isosceles triangle, the median on the base is also the altitude on the base.

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Solution

In isosceles $Δ A B C$ , Let $A B = A C$ and AD is the median then
$\begin{matrix} - - \to A D = \frac{1}{2} (- - \to A B + - - \to A C) a n d - - \to B C = - - \to B A + - - \to A C H e r e, - - \to D A \cdot - - \to D B = - - - \to A D \cdot \frac{1}{2} - - \to C B = - - - \to A D \cdot \frac{1}{2} - - \to B C = \frac{1}{2} - - \to A D \cdot - - \to B C = \frac{1}{4} (- - \to A B + - - \to A C) \cdot (- - \to B A + - - \to A C) = \frac{1}{4} (- - \to A B + - - \to A C) \cdot (- - \to A C - - - \to A B) = \frac{1}{4} (- - \to A C \cdot - - \to A C) - (- - \to A B + - - \to A B) = \frac{1}{4} (A C^{2} - A B^{2}) = \frac{1}{4} \times 0 = 0 i . e . ∠ A D B = 90^{\circ} \end{matrix}$
So, $A D$ is perpendicular to $B C$ .

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