Question

If the median to the base of a triangle is perpendicular to the base, then triangle is isosceles. 

Answer 1

 

Let ∆ABC be a triangle such that AD is the median. Taking A as the origin, let the position vectors of B and C be \[\vec{b}\] and \[\vec{c}\] respectively.

 Then, 

Position vector of D = \[\frac{\vec{b} + \vec{c}}{2}\].....................(Mid-point formula) 

Now, 

\[\vec{AD}\] = Position vector of D − Position vector of A =\[\frac{\vec{b} + \vec{c}}{2}\] 

\[\vec{BC}\] = Position vector of C − Position vector of B= \[\vec{c} - \vec{b}\] 

Since 

\[\vec{AD} \perp \vec{BC}\] 

\[\therefore \vec{AD} . \vec{BC} = 0\]
\[ \Rightarrow \frac{1}{2}\left( \vec{b} + \vec{c} \right) . \left( \vec{c} - \vec{b} \right) = 0\]
\[ \Rightarrow \left( \vec{c} + \vec{b} \right) . \left( \vec{c} - \vec{b} \right) = 0\]
\[ \Rightarrow \left| \vec{c} \right|^2 - \left| \vec{b} \right|^2 = 0\]
\[ \Rightarrow \left| \vec{c} \right| = \left| \vec{b} \right|\]
\[ \Rightarrow AC = AB \] 

Hence, the ∆ABC is an isosceles triangle.