In the adjacent figure, ABCD is a parallelogram and line segments AE and CF bisect the angles A and C respectively.

Solution:

Refer to the figure of the book.

$\angle \mathrm{A}=\angle \mathrm{C}$         (opposite angles of $a$ parallelogram are equal)

$\Rightarrow \frac{1}{2} \angle \mathrm{A}=\frac{1}{2} \angle \mathrm{C}$

$=>\angle \mathrm{EAD}=\angle \mathrm{FCB}$     ( $A E$ and $C F$ bisect the angles $A$ and $C$, respectively)

In $\Delta A D E$ and $\Delta C B F$ :

$\angle B=\angle D \quad$ (o pposite angles of a parallelogram are equal)

$\angle E A D=\angle F C B \quad$ ( $p$ roved above)

$A D=B C \quad(o$ pposite sides of a parallelogram are equal $)$

By AAS concruency criteria :

$\Delta A D E \cong \Delta B C F$

$D E=B F$   (corresponding parts of congruent triangles)

$C D=A B \quad$ (opposite sides of a parallelogram are equal)

Also, $C D-D E=A B-B F$

$\Rightarrow C E=A F$

$A B C D$ is a paralleleogram.

$\therefore C D \| A B \quad$ (opposite sides of a parallelogram are parallel)

$=>C E \| A F$

If one pair of sides of a quadrilateral is parallel and equal, then it is a parallelogram.

Therefore, AECF is a parallelogram.

$\therefore A E \| C F$