Question

prove that the sum of the vectors directed from the vertices to the mid point?


pls help me frnds​

Answer 1

Step-by-step explanation:

Let the three vertices of the triangle have location vectors aa, bb, and cc. Then the midpoints of the sides are (a+b)/2(a+b)/2, (b+c)/2(b+c)/2, and (a+c)/2(a+c)/2. The three vectors from vertex to midpoint are

a+b2−c,b+c2−a,a+c2−b,a+b2−c,b+c2−a,a+c2−b,

and these add up to zero.

Note. Nowhere does this use the fact that in order to have a triangle, the points with location vectors aa, bb, and cc must be non-collinear. So the results are true for any three distinct points in any vector space (given the appropriate interpretation of “vertex to midpoints of opposite sides”) and not just for a triangle in R2R2.

OR

Here’s a visual, geometric proof. Start with a triangle and its three medians.

We’re considering the sum of the three vectors

 AD−→−,AD→, BE−→−,BE→, and CF−→−.CF→.

One thing we know about vectors going around a triangle is that their sum is zero:

 AB−→−+BC−→−+CA−→−=0.AB→+BC→+CA→=0.

So half that sum is also zero:

 AF−→−+BD−→−+CE−→−=0.AF→+BD→+CE→=0.

But 

AF−→−=AG−→−+GF−→−,AF→=AG→+GF→, BD−→−=BG−→−+GD−→−,BD→=BG→+GD→, and CE−→−=CG−→−+GE−→−.CE→=CG→+GE→.

Therefore, AD−→−+BE−→−+CF−→−=0.

Refer the attachment above for

visual geometric

proof..

Hope this might help you..

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