Question

In the quadrilateral $ABCD$. If $AB=BC$ and $AD=DC$, find the measure of $\angle BCD$.
$\left[5\right]$

Answer 1

Join $BD$
Compare triangle $ABD$ and $BCD$
$AB=BC$
$AD=CD$, and
$BD$ is common
So, $\Delta ABD\cong \Delta BCD$
$\left[1\right]$
Clearly $BD$ will bisect angles $\angle ABC$ and $\angle ADC$
$\left[1\right]$

$\angle DBC={54}^o,\angle BDC={21}^o$
$\left[1\right]$
$\therefore \angle BCD={180}^o-({54}^o+{21}^o)$ [$\because$ using angle sum property]
$\angle BCD={180}^o-{75}^o$
$\angle BCD={105}^o$
$\left[2\right]$