Question

Prove that in a rhombus with angle of 60$°$, the shorter diagonal divides the rhombus into two equilateral triangles.

Answer 1

Consider the following figure.


$$\begin{gathered} \angle \mathrm{ADC}=60°\left(\mathrm{Given}\right) \\ \angle \mathrm{ABC}=60°(\mathrm{Opposite}\mathrm{angles}\mathrm{of}\mathrm{a}\mathrm{rhombus}\mathrm{are}\mathrm{equal}) \\ \mathrm{Now},\mathrm{consider}△\mathrm{ADC} \\ \because \mathrm{AD}=\mathrm{CD}(\mathrm{Sides}\mathrm{of}\mathrm{a}\mathrm{rhombus}) \\ \because \angle \mathrm{DAC}=\angle \mathrm{DCA}\left(\mathrm{Angles}\mathrm{opposite}\mathrm{to}\mathrm{equal}\mathrm{sides}\mathrm{are}\mathrm{equal}\right) \\ \angle \mathrm{ADC}+\angle \mathrm{DCA}+\angle \mathrm{CAD}=180°(\mathrm{Angle}\mathrm{sum}\mathrm{property}) \\ \Rightarrow 2\angle \mathrm{DCA}+60°=180°[\mathrm{From}(\mathrm{i}\left)\right] \\ \Rightarrow \angle \mathrm{DCA}=\angle \mathrm{DAC}=60° \\ \mathrm{Similarly}, \\ \angle \mathrm{BAC}=\angle \mathrm{BCA}=60° \\ \mathrm{AC},\mathrm{which}\mathrm{is}\mathrm{the}\mathrm{shorter}\mathrm{diagonal},\mathrm{divides}\mathrm{the}\mathrm{rhombus}\mathrm{into}\mathrm{two}\mathrm{equilateral}\mathrm{triangles}:△\mathrm{ADC}\mathrm{and}△\mathrm{ABC}. \end{gathered}$$