The triangle midsegment theorem
- Thread starter Faye
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Let's plot those points and join them as follows:
Now, one side of the triangle will have a side that passes through point \(K\) and will be parallel to segment \(\overline{JL}\), and twice as long as this segment, with \(K\) at its midpoint. Can you find two of the vertices of the triangle?
Now, one side of the triangle will have a side that passes through point \(K\) and will be parallel to segment \(\overline{JL}\), and twice as long as this segment, with \(K\) at its midpoint. Can you find two of the vertices of the triangle?
pka
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Here completely different way to go.
Say the vertices of the triangle are : \(\displaystyle A:\,(x_A,y_A),\,B:\,(x_B,y_B),~\&~C:\,(x_C,y_C)\)
\(\displaystyle \begin{cases}\dfrac{x_A+x_B}{2}=-2 &,\quad \dfrac{y_A+y_B}{2}=0 \\\dfrac{x_A+x_C}{2}=0 &,\quad \dfrac{y_A+y_C}{2}=2 \\\dfrac{x_C+x_B}{2}=-4 &,\quad \dfrac{y_C+y_B}{2}=1 \\\end{cases}\)
At first it appears that there are too many variables.
But note that \(\displaystyle y_A=-y_B~\&~x_A=-x_C\).
Thank you!
How did you plot those other points? Slope?
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Yes, each side of the triangle is parallel to the side opposite the vertex of the inner triangle through which it passes, and is twice as long.
Great! Thanks again!