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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