Question

ABCD is a quadrilateral in which$ \mathrm{AD}=\mathrm{BC}$ and $ \angle \mathrm{DAB}=\angle \mathrm{CBA}$ .

Prove that $ \left(\mathrm{i}\right)△\mathrm{ABD}\cong △\mathrm{BAC}$

$ \left(\mathrm{ii}\right) \mathrm{BD} = \mathrm{AC}$

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

Answer 1

Step $1:$ Proving $△\mathrm{ABD}\cong △\mathrm{BAC}$:

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

$\begin{array}{rcl}\mathrm{AD}& =& \mathrm{BC}\left(\mathrm{Given}\right)\\ \angle \mathrm{DAB}& =& \angle \mathrm{CBA}\left(\mathrm{Given}\right)\\ \mathrm{AB}& =& \mathrm{BA}(\mathrm{Common}\mathrm{side})\end{array}$

Therefore, By SAS congruence rule

$\therefore △\mathrm{ABD}\cong △\mathrm{BAC}$

Step $2:$ Proving $\mathrm{BD}=\mathrm{AC}$ and $\angle \mathrm{ABD}=\angle \mathrm{BAC}$:

BD and AC will be equal as they are corresponding parts of congruent triangles(CPCT)

$\therefore \mathrm{BD}=\mathrm{AC}$

$\angle \mathrm{ABD}$ and $\angle \mathrm{BAC}$ will be equal as they are corresponding parts of congruent triangles(CPCT).

$\therefore \angle \mathrm{ABD}=\angle \mathrm{BAC}$

Hence, proved