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Vector Algebra

Question
CBSEENMA12034150

Prove that the median to the base of isosceles triangle is perpendicular to the base.

Solution
Let ABC be an isosceles triangle with AB = AC and let AD be the median to the base BC. Then D is the middle point of BC.

Take A as origin. Let straight b with rightwards arrow on top comma space straight c with rightwards arrow on top be position vectors of B and C respectively with respect to origin A such that AB with rightwards arrow on top space equals space straight b with rightwards arrow on top comma space space AC with rightwards arrow on top space equals straight c with rightwards arrow on top.  Then position vector of D w.r.t A is fraction numerator straight b with rightwards arrow on top plus straight c with rightwards arrow on top over denominator 2 end fraction.
Now,    BC with rightwards arrow on top space equals space straight P. straight V. space of space straight C space minus space straight P. straight V. space of space straight B space equals straight c with rightwards arrow on top minus straight b with rightwards arrow on top
therefore space space space BC with rightwards arrow on top. space AD with rightwards arrow on top space equals space left parenthesis straight c with rightwards arrow on top space minus straight b with rightwards arrow on top right parenthesis. space open parentheses fraction numerator straight b with rightwards arrow on top plus straight c with rightwards arrow on top over denominator 2 end fraction close parentheses space equals space 1 half left parenthesis straight c with rightwards arrow on top space minus space straight b with rightwards arrow on top right parenthesis space left parenthesis straight c with rightwards arrow on top plus straight b with rightwards arrow on top right parenthesis
space space space space space space space space space space space space space space space space space space space space space space equals space 1 half left parenthesis straight c with rightwards arrow on top squared space minus space straight b with rightwards arrow on top squared right parenthesis space equals space 1 half left parenthesis open vertical bar straight c with rightwards arrow on top close vertical bar squared space minus space open vertical bar straight b with rightwards arrow on top close vertical bar squared right parenthesis
space space space space space space space space space space space space space space space space space space space space space space space equals space 1 half left parenthesis AC squared minus space AB squared right parenthesis space equals space 1 half left parenthesis 0 right parenthesis space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets because AC space equals space AB close square brackets
therefore space space space BC with rightwards arrow on top space is space perpendicular space to space AD with rightwards arrow on top.
Hence space the space result.