Question

In the given figure $ABCD$ is a square. Find the measure of $\angle CAD$.

Answer 1

In $\Delta ADC$
$DA=DC$ (all sides of a square are equal)
$\Rightarrow \angle ACD=\angle CAD$ [Angle opposite to the equal sides are equal]
Let $\angle ACD=\angle CAD=x^{\circ }$
$\Rightarrow x+x+90=180$
Since the sum of the angles of a triangle is ${180}^{\circ }$
$\Rightarrow 2x+90=180$
$\Rightarrow 2x=90$
$\Rightarrow x=\frac{90}{2}$
$\therefore x=\angle CAD={45}^{\circ }$