$A B C D$ is a quadrilateral in which $A D=B C$ and $\angle D A B=\angle C B A$ (See the given figure). Prove that

(i) $\triangle \mathrm{ABD} \cong \triangle \mathrm{BAC}$

(ii) $B D=A C$

(iii) $\angle \mathrm{ABD}=\angle \mathrm{BAC}$.

Solution:

In $\triangle \mathrm{ABD}$ and $\triangle \mathrm{BAC}$,

$\mathrm{AD}=\mathrm{BC}$ (Given)

$\angle \mathrm{DAB}=\angle \mathrm{CBA}$ (Given)

$\mathrm{AB}=\mathrm{BA}$ (Common)

$\therefore \triangle \mathrm{ABD} \cong \triangle \mathrm{BAC}(\mathrm{By} \mathrm{SAS}$ congruence rule $)$

$\therefore \mathrm{BD}=\mathrm{AC}(\mathrm{By} \mathrm{CPCT})$

And, $\angle \mathrm{ABD}=\angle \mathrm{BAC}(\mathrm{By} \mathrm{CPCT})$
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