Question

In the figure $ABCD,BD=BC=AD$ and $\angle ACD={37}^{\circ }$. find $\angle ADB$

Answer 1

Since, $BD=BC=AD$

Hence, $\angle BDC=\angle BCD=\angle ACD={37}^{\circ }$

$\angle ACD=\angle ABD={37}^{\circ }$ (Opposite angles of $AD$)

$\angle ADB=y$

In $△BCD,$

$\angle BCD+\angle CDB+\angle DBC={180}^{\circ }$

${37}^{\circ }+{37}^{\circ }+\angle DBC={180}^{\circ }$

$\angle DBC={106}^{\circ }$

Now,

$\angle DBC={106}^{\circ }=\angle DAC=\angle BAD$ (Opposite angles of side $CD$)

In $△ABD$,

$AD=BD$

$\angle ABD=\angle BAD={37}^{\circ }$

$\angle BAD+\angle ADB+\angle DBA={180}^{\circ }$

${106}^{\circ }+{37}^{\circ }+y={180}^{\circ }$

$y={37}^{\circ }$

This is the required solution.