Rhombus in two equilateral triangle-by Dao Thanh Oai
$MN$ and $QP$ parallel and equal $\frac{A_1B_1}{2}$ (1)
$MQ$ and $NP$ parallel and equal $\frac{A_2B_2}{2}$ (2)
$\Delta OA_1B_1=\Delta OA_2B_2$ => $A_1B_1=A_2B_2$ (3)
Since (1) (2) and (3) => MNPQ is rhombus
$A_1O$meet $A_2B_2$at D; $A_1B_1$ meet $A_2B_2$ at C.
$\angle ODA_2=60^0+\angle DA_2O=\angle DA_1C+\angle A_1CD$ (4)
$\angle DA_2O=\angle DA_1C$ (5)
Since (4) and (5) => $ A_1CD=60^0$ => $ \angle MNP=60^0$ => $ MNPQ$ is rhombus, the rhombus consists two equilateral triangle
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