Show that each diagonal of a rhombus bisects the angle through which it passes.
In $\triangle \mathrm{AED}$ and $\Delta \mathrm{DEC}:$
$\mathrm{AE}=\mathrm{EC}$ (diagonals bisect each other)
$\mathrm{AD}=\mathrm{DC}$ (sides are equal)
$\mathrm{DE}=\mathrm{DE}$ (common)
$\mathrm{By} \mathrm{SSS}$ congruence :
$\triangle \mathrm{AED} \cong \Delta \mathrm{CED}$
$\angle \mathrm{ADE}=\angle \mathrm{CDE}$ (c. p.c.t)
Similarly, we can prove $\Delta \mathrm{AEB}$ and $\Delta \mathrm{BEC}, \Delta \mathrm{BEC}$ and $\Delta \mathrm{DEC}, \Delta \mathrm{AED}$ and $\Delta \mathrm{AEB}$ are congruent to each other.
Hence, diagonal of a rhombus bisects the angle through which it passes.
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