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