For this one complex numbers should work just fine. Let E be the center of the equilateral triangle constructed on AB. F, G, H are the same for sides BC, CD, DA respectively. Let small letters (like a,b,c etc) be the complex numbers that represent the points with the correspondinc\g s\capital letters (A, B, C etc.) Let z=cos 120+i*sin 120. Then e=(a-bz)/(1-z), f=(b-cz)/(1-z), g=(c-dz)/(1-z), h=(d-az)/(1-z). If we calculate A=(e-g)/(h-f) and its conjugate and we take into account the fact that |a-c|=|b-d| we get A {conjugate} = -A, which is equivalent to A=(e-g)/(h-f) = k*i with k a real number and i=(-1)^1/2 (*). But (*) is the condition for the lines EG and FH to be perpendicular. This is the usual method by which these prbs involving equilateral triangles can be solved (I think there are other methods, but for this specific problem this the best way I could find). [Moderator edit: Also discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=17861 ]

Let $ ABCD$ be a convex quadrilateral. We erect four equilateral triangles $ WAB$, $ XBC$, $ YCD$, $ ZDA$ on its sides $ AB$, $ BC$, $ CD$, $ DA$ exterior to the quadrilateral $ ABCD$, and denote the centroids of these triangles by $ S_{1}$, $ S_{2}$, $ S_{3}$, $ S_{4}$, respectively. Prove that: $ S_{1}S_{3}$ is perpendicular to $ S_{2}S_{4}$ if and only if $ AC = BD$. Solution by Myth: It is well known that $ S_1S_3\perp S_2S_4$ iff $ S_1S_2^2 + S_3S_4^2 = S_2S_3^2 + S_4S_1^2$. Apply cosine theorem to $ AS_1S_4$ and $ CS_2S_3$. We obtain: \[ S_4S_3^1 + S_2S_3^2 = \frac {1}{3}(AB^2 + BC^2 + CD^2 + DA^2) - \frac {2}{3}AB\cdot AD\cdot\cos(A + \pi/3) - \frac {2}{3}CB\cdot CD\cdot\cos(C + \pi/3). \] We can write \[ BD^2 = AB^2 + AD^2 - 2AB\cdot AD\cdot\cos(A) = CB^2 + CD^2 - 2CB\cdot CD\cdot\cos(C) \] and \[ AB\cdot AD\cdot\sin(A) + CB\cdot CD\cdot\sin(C) = 2S_{ABCD}. \] So after some easy transformation we conclude that \[ S_4S_3^1 + S_2S_3^2 = \frac {1}{6}(AB^2 + BC^2 + CD^2 + DA^2) + \frac {1}{3}BD^2 + \frac {\sqrt {3}}{6}S_{ABCD}. \] Analogously, \[ S_4S_1^2 + S_2S_3^2 = \frac {1}{6}(AB^2 + BC^2 + CD^2 + DA^2) + \frac {1}{3}AC^2 + \frac {\sqrt {3}}{6}S_{ABCD}. \] Omid Hatami notes: Also this can easily proved with vector geometry.

Could anyone post the vector solution ?

Let $ABCD$ be a quadrilateral with equilateral triangles $ABE,BFC,CGD,DHA$ constructed externally.Then the coordinates of $E,F,G,H$ are $e=b+(a-b)\epsilon$ $f=c+(b-c)\epsilon$ $g=d+(c-d)\epsilon$ $h=a+(d-a)\epsilon$ Thus coordinates of $O_1,O_2,O_3,O_4$ are $o_1=\frac{a+2b+(a-b)\epsilon}{3}$ $o_2=\frac{b+2c+(b-c)\epsilon}{3}$ $o_3=\frac{c+2d+(c-d)\epsilon}{3}$ $o_4=\frac{d+2a+(d-a)\epsilon}{3}$ where $\epsilon=cos60^{\circ}+sin60^{\circ}$ After a bit of manipulation using the relations we get that $|o_1-o_2|^2=|o_2-o_3|^2=|o_3-o_4|^2=|o_4-o_1|^2$ and $o_1-o_3|^2=|o_2-o_4|^2 \Leftrightarrow |a-c|^2=|b-d|^2$(It won't be difficult at all if we use the realtion $|a-b|^2=(a-b)(\overline{a}-\overline{b})$).So we are through.... Another related problem: On the sides of $AB,BC,CD,DA$ of quadrilateral $ABCD$ and exterior to the quadrilateral,we construct squares of centers $O_1,O_2,O_3,O_4$ respectively.Show that $O_1O_3 \perp O_2O_4$ and $O_1O_3=O_2O_4$

Haha this problem is crying for complex bash... Let $a,b,c,d,o_1,o_2,o_3,o_4$ be the affixes of $A,B,C,D,O_1,O_2,O_3,O_4$, respectively. Note that $a$ is the result of rotating $b$ by $\tfrac{\pi}{3}$ about $o_1$. We have the equation \[ a-o_1=\varepsilon(b-o_1) \Rightarrow o_1=\dfrac{a-b\varepsilon}{1-\varepsilon} \] where $\varepsilon=e^{\tfrac{2\pi i}{3}}$. Similarly, $o_2=\tfrac{b-c\varepsilon}{1-\varepsilon}, o_3=\tfrac{c-d\varepsilon}{1-\varepsilon}, o_4=\tfrac{d-a\varepsilon}{1-\varepsilon}$. We want to show $O_1O_3 \perp O_2O_4$, which from Orthogonality criterion is equivalent to \[ \dfrac{\dfrac{a-b\varepsilon}{1-\varepsilon}-\dfrac{c-d\varepsilon}{1-\varepsilon}}{\overline{\dfrac{a-b\varepsilon}{1-\varepsilon}}-\overline{\dfrac{c-d\varepsilon}{1-\varepsilon}}} = -\dfrac{\dfrac{b-c\varepsilon}{1-\varepsilon}-\dfrac{d-a\varepsilon}{1-\varepsilon}}{\overline{\dfrac{b-c\varepsilon}{1-\varepsilon}}-\overline{\dfrac{d-a\varepsilon}{1-\varepsilon}}} \] \[ \iff \dfrac{a-c-(b-d)\varepsilon}{\overline{a-c}-\overline{(b-d)\varepsilon}} = - \dfrac{b-d-(c-a)\varepsilon}{\overline{b-d}-\overline{(c-a)\varepsilon}}\] \[ \iff \left[ a-c-(b-d)\varepsilon \right] \left[ \overline{b-d}-\overline{(c-a)\varepsilon} \right] = -\left[ b-d-(c-a)\varepsilon \right] \left[ \overline{a-c}-\overline{(b-d)\varepsilon} \right] \] \[ \iff (a-c)\overline{b-d}-(a-c)\overline{(c-a)}\overline{\varepsilon}-(b-d)\overline{b-d}\varepsilon+(b-d)\overline{c-a}\varepsilon\overline{\varepsilon}=-(b-d)\overline{a-c}+(b-d)\overline{(b-d)}\overline{\varepsilon}+(c-a)\overline{a-c}\varepsilon-(c-a)\overline{b-d}\varepsilon\overline{\varepsilon} \] Let $x=a-c$, $y=b-d$. (Should've done this earlier but whatever...) \[ \iff x\overline{y}+x\overline{x}\dfrac{1}{\varepsilon}-y\overline{y}\varepsilon-y\overline{x} = -y\overline{x}+y\overline{y}\dfrac{1}{\varepsilon}-x\overline{x}\varepsilon+x\overline{y}\] \[\iff x\overline{x}(\varepsilon+\dfrac{1}{\varepsilon})-y\overline{y}(\varepsilon+\dfrac{1}{\varepsilon})=0\] \[\iff x\overline{x}=y\overline{y} \] \[\iff AC=BD \] which is given.

Really if is only if, like said the post #3 and this true for internal equilaterals too. This is consequence of the area $\Delta '$ in the Napoleon triangle of $\triangle ABC$ $\Delta' = \frac{[ABC]}{2}+\frac{\sqrt{3}(a^2+b^2+c^2)}{24}$ (outer) $\Delta'=\frac{a^2+b^2+c^2-4\sqrt{3}[ABC]}{6}$ (internal, this implies Weitzenböck's inequality) Is sufficient prove that $O_1O_2^2+O_3O_4^2=O_2O_3^2+O_1O_4^2$ and this trivial, with this formule because area of a equilateral triangle is $\frac{l^2\sqrt{3}}{4}$

Let $\omega$ be a third root of unity. Then we can compute the centers are $\frac{a\omega - b}{\omega-1}$ and cyclic permutations, so it suffices to show that$$\frac{o_1-o_3}{o_2-o_4} = \frac{\omega(a-c)-(b-d)}{(b-d)\omega + (a-c)}$$is pure imaginary. The conjugate equals \begin{align*} \frac{\overline \omega \overline{a-c} - \overline{b-d}}{\overline{b-d} \overline \omega + \overline{a-c}} &= \frac{\overline{a-c} - \overline{b-d} \omega}{\overline{b-d}-\overline{a-c} \omega} \\ &= \frac{\frac{\overline{b-d}(b-d)}{a-c} - \frac{\overline{a-c}(a-c)}{b-d} \omega}{\frac{\overline{a-c}(a-c)}{b-d} + \frac{\overline{b-d}(b-d)}{a-c}\omega} \\ &= \frac{(b-d) - (a-c) \omega}{(b-d) \omega + a-c}, \end{align*}which is the negative of the original, as needed.