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# ABCD is a quadrilateral in which AD = BC and $$\angle{DAB}$$ = $$\angle{CBA}$$ (see figure). Prove that (i) $$\triangle{ABD}$$ $$\displaystyle \cong$$ $$\triangle{BAC}$$ (ii) BD = AC (iii) $$\angle{ABD}$$ = $$\angle{BAC}$$

In $$\triangle{ABC}$$ and $$\triangle{BAC}$$, we have

Also,
$$\angle{DAB}$$ = $$\angle{CBA}$$ ...(Given)

And AB is a common side.

Therefore, $$\triangle{ABD}$$ $$\displaystyle \cong$$ $$\triangle{BAC}$$ ...(SAS congruency test)

Hence, BD = AC ...(By CPCT)

and $$\angle{ABD}$$ = $$\angle{BAC}$$ ...(By CPCT)
Hence, proved.