0
$\begingroup$

Consider triangle $ABC$ We construct $A',B',C'$ like the picture below:

enter image description here

(Here the name of points are different.)$A,B,C$ are mid points.Prove that the intersection of medians is the same point in both triangles.

What should we do to prove that?The medians are not the same so working with the definition of all medians seems to be useless so I considered one median and found a point on it that divide it into two part that one part is two times bigger than the other but this definition also didn't work.What should I do now?

$\endgroup$
1
$\begingroup$

We have $D=2A-B$, $E=2C-A$ and $F=2B-C$. Therefore, the intersections of medians of the triangle $DEF$ is $$ \frac{D+E+F}{3} = \frac{2A-B+2C-A+2B-C}{3} = \frac{A+B+C}{3} $$ which is the intersection of medians of the triangle $ABC$

$\endgroup$
  • $\begingroup$ You mean the coordinate?Why do we have so? $\endgroup$ – Taha Akbari Jul 11 '17 at 13:24
  • $\begingroup$ It is not required to have the actual coordinates. It is only important to know that the figure could be embedded into a Cartesian coordinate system. The proof works independently of the particular coordinates. $\endgroup$ – Reinhard Meier Jul 11 '17 at 14:24
0
$\begingroup$

$$A+B+C=\frac{D+B}{2}+\frac{F+C}{2}+\frac{A+E}{2}$$ or $$A+B+C=D+F+E$$ or $$\frac{A+B+C}{3}=\frac{D+F+E}{3}$$ and we are done!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.