In the given figure ABCD is a square.

Refer to the figure given in the book.

In $\Delta A D C:$

$D A=D C$   (all sides of a square are equal)

$\Rightarrow \angle A C D=\angle C A D$

Let $\angle A C D=\angle C A D=x^{\circ} \quad$ [Angle opposite to the equal sides are equal]

$x+x+90=180 \quad$ [since the sum of the angles of a triangle is $180^{\circ}$ ]

$\Rightarrow 2 x+90=180$

$\Rightarrow 2 x=90$

$\Rightarrow x=\frac{90}{2}$

$\Rightarrow x=45$

$\therefore \angle C A D=45^{\circ}$