In the adjacent figure, ABCD is a parallelogram and line segments AE and CF bisect the angles A and C respectively.
In the adjacent figure, ABCD is a parallelogram and line segments AE and CF bisect the angles A and C respectively. Show that AE||CF.
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.
$\therefore A E \| C F$
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