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Answer :

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.

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