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# 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,
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)
$$\therefore$$ $$\triangle{BDC}$$ $$\displaystyle \cong$$ $$\triangle{BDA}$$ ...(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 $$\angle{D}$$.