Question
In a rhombus $ABCD$ show that diagonal $AC$ bisects $\angle\text{A}$ as well as $\angle\text{C}$ and diagonal $BD$ bisects $\angle\text{B}$ as well as $\angle\text{D}.$
✓
Answer
In $\triangle\text{ABC}$ and $\triangle\text{ADC},$
$\text{AB = AD}$ (sides of a rhombus are equal)
$\text{BC = CD}$ (sides of a rhombus are equal)
$\text{AC = AC}$ (common)
$\therefore\triangle\text{ABC}\cong\triangle\text{ADC}$ (by $SSS$ congruence criterion)
$\Rightarrow\angle\text{BAC}=\angle\text{DAC}$ and $\angle\text{BCA}=\angle\text{DCA}$ $(C.P.C.T.)$
$\Rightarrow\text{AC}$ bisects $\angle\text{A}$ as well as $\angle\text{C}.$
Similarly,
In $\triangle\text{BAD}$ and $\triangle\text{BCD},$
$\text{AB = BC}$ (sides of a rhombus are equal)
$\text{AD = CD}$ (sides of a rhombus are equal)
$\text{BD = BD}$ (common)
$\therefore\triangle\text{BAD}\cong\triangle\text{BCD}$ (by $SSS$ congruence criterion)
$\Rightarrow\angle\text{ABD}=\angle\text{CBD}$ and $\angle\text{ADB}=\angle\text{CDB}$ $(C.P.C.T.)$
$\Rightarrow\text{BD}$ bisects $\angle\text{B}$ as well as $\angle\text{D}.$
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