ABCD is a rhombus. Show that diagonal AC bisects \(\angle{A}\) as well as \(\angle{C}\) and diagonal BD bisects \(\angle{B}\) as well as \(\angle{D}\).

Given: ABCD is a rhombus.
i.e., AD = AB = BC = CD ... (i)

To prove: (i) Diagonal AC bisect \(\angle{A}\) as well as \(\angle{C}\).
(ii) Diagonal BD bisects \(\angle{B}\) as well as \(\angle{D}\).

Proof: Let AC and BD are the diagonals of rhombus ABCD.
In \(\triangle{ABC}\) and \(\triangle{ADC}\), we have,
AD = AB ...(Given)
AC = CA ...(Common side)
CD = BC ...(From eq. (i))
\(\therefore \) \(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{ADC}\) ...(By SSS rule)
Thus, \(\angle{DAC}\) = \(\angle{BAC}\) ...(By CPCT)
Also, \(\angle{DCA}\) = \(\angle{BCA}\)
Also, \(\angle{DAC}\) = \(\angle{DCA}\)
And \(\angle{BAC}\) = \(\angle{BCA}\)
This shows that, Diagonal AC bisect \(\angle{A}\) as well as \(\angle{C}\).

Now, in \(\triangle{BDC}\) and \(\triangle{BDA}\), we have,
AB = BC ...(Given)
BD = BD ...(Common side)
AD = CD ...(Given)
\(\therefore \) \(\triangle{BDC}\) \(\displaystyle \cong \) \(\triangle{BDA}\) ...(By SSS rule)
Thus, \(\angle{BDA}\) = \(\angle{BDC}\) ...(By CPCT)
Also, \(\angle{DBA}\) = \(\angle{DBC}\)
Also, \(\angle{BDA}\) = \(\angle{DBA}\)
And \(\angle{BDC}\) = \(\angle{DBC}\)
This shows that, Diagonal BD bisects \(\angle{B}\) as well as \(\angle{D}\).
Hence, proved.