In the figure, ABCD is a trapezium and BCEF is a square. $\angle DCE = 203^\circ$.
Find $\angle ABC$. Ans: ___ $^\circ$

1. Find $\angle BCD$ using angles at a point

$\angle BCE = 90^\circ$ (interior angle of a square)
$\angle BCD + \angle BCE + \text{reflex } \angle DCE = 360^\circ$ (angles at a point)
$\angle BCD + 90^\circ + 203^\circ = 360^\circ$
$\angle BCD = 360^\circ - 293^\circ = 67^\circ$

2. Find $\angle ABC$ using interior angles between parallel lines

Since $AB \parallel DC$ (ABCD is a trapezium),
$\angle ABC + \angle BCD = 180^\circ$ (interior angles)
$\angle ABC = 180^\circ - 67^\circ = 113^\circ$