Verify that the following 3 points (0,0), (a,0) and
(a/2, [sqrt{3}a]/2) are the vertices of an equilateral triangle. Then show that the midpoints of the 3 sides are vertices of a second equilateral triangle.
Calculate the distance between the three pairs of vertices. Hopefully you get the same answer each time ....
Find the coordinates of the midpoint between each of the three pairs of vertices. Now calculate the distance between the three pairs of these vertices. Hopefully you get the same answer each time ....
The distance formula from (a,0) to (a/2,a/2(3^.5))
is ((a-a/2)^2 +(0-a/2(3^.5))^2).5 and =((a^2)/4 + 3a^2/4)^2=(4a^2/4)=a
similarly for the other two sides.
A standard theorem states that the line segment joining the midpoints of two
sides of a triangle is parallel to the third side and is one-half its length. Therefore the three sides of the embedded triangle are equal , so it is an equilateral triangle also,
Your notation is a little tough to read. Allow me:
$\displaystyle A(0,0), B(a,0), C\left(\frac{a}{2},\frac{\sqrt{3}a}{2}\right)$
$\displaystyle d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
$\displaystyle {\color{red}AB}=\sqrt{(a-0)^2+(0-0)^2}=\sqrt{a^2}={\color{red}a}$
$\displaystyle {\color{red}AC}=\sqrt{\left(\frac{a}{2}-0\right)^2+\left(\frac{\sqrt{3}a}{2}-0\right)^2}=\sqrt{\frac{a^2}{4}+\frac{3a^2}{4}}=\s qrt{\frac{4a^2}{4}}=\sqrt{a^2}={\color{red}a}$
$\displaystyle {\color{red}BC}=\sqrt{\left(\frac{a}{2}-a\right)^2+\left(\frac{\sqrt{3}a}{2}-0\right)^2}=\sqrt{\left(\frac{a-2a}{2}\right)^2+\frac{3a^2}{4}}=$
$\displaystyle \sqrt{\left(\frac{-a}{2}\right)^2+\frac{3a^2}{4}}=\sqrt{\frac{a^2}{4} +\frac{3a^2}{4}}=\sqrt{\frac{4a^2}{4}}=\sqrt{a^2}= {\color{red}a}$
$\displaystyle {\color{red}AB=AC=BC}$
Tim, your explanation about the embedded triangle was "right on".