Question
In quadrilateral $\text{ABCD, AD = BC}$ and $\ce{BD = CA}.$ Prove that: $(i)\angle ADB =\angle BCA$ $(ii)\angle DAB =\angle CBA$
✓
Answer
Given: In quadrilateral $\mathrm{ABCD}, \mathrm{AD}=\mathrm{BC}$ and $\mathrm{BD}=\mathrm{AC}$.
To Prove:
$(i) \angle \mathrm{ADB}=\angle \mathrm{BCA}$
$(ii) \angle \mathrm{DAB}=\angle \mathrm{CBA}$
Proof:
In $\triangle \mathrm{ABD}$ and $\triangle \mathrm{BAC}$,
$A D=B C ....($given$)$
$B D=C A ....($given$)$
$A B=A B ....($common$)$
$\therefore \triangle \mathrm{ABD} \cong \triangle \mathrm{BAC} \ldots. ($by $\text{SSS}$ congruence criterion$)$
$\angle \mathrm{ADB}=\angle \mathrm{BCA}$
$\angle \mathrm{DAB}=\angle \mathrm{CBA}$
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