in triangle ABC if AD is the median show that AB sq.+ Ac sq.2=2ADsq. + 2 BD square
Answers
Step-by-step explanation:
This is made particularly easy with vectors.
Put A at the origin. To avoid confusion between vectors and points, let's use lower case for vectors, but otherwise the same letters. So A at the origin means that a = 0. Also the vector along the side AB is b. Similarly, the side AC corresponds to c, and the median AD corresponds to d.
Also, we have d = ( b + c ) / 2, since D is the midpoint of BC. So the vector along BD corresponds to d - b = ( b + c ) / 2 - b = ( c - b ) / 2.
The square of the length of a vector is just the dot product of the vector with itself. So the right hand side of the equation in the problem is
2 AD² + 2 BD²
= 2 d · d + 2 ( d - b ) · ( d - b )
= ( ( b + c ) · ( b + c ) + ( c - b ) · ( c - b ) ) / 2
= ( b · b + 2 b · c + c · c + c · c - 2 b · c + b · b ) / 2
= b · b + c · c
= AB² + AC²