Question

In the adjacent figure, $\text{ABCD}$ is a parallelogram and line segments $\text{AX}$ and $\text{CY}$ bisect the angles $\text{A}$ and $\text{C}$ respectively. Show that $\text{AX}\parallel \text{CY}.$

Answer 1

Given: $ABCD$ is Parallelogram

To prove: $\text{AX}||\text{CY}$

Proof:

$\angle A=\angle C$ $[$opposite angles of parallelogram$]$

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

$\therefore \angle 1=\angle 2......\left(1\right)$ $[AX$ is the bisector of $\angle A$ and $CY$ is the bisector of $\angle C]$

Now, $AB||CD$ and $CY$ is a transversal.

$\therefore \angle 2=\angle 3......\left(2\right)$ [Alternate angle are equal]

From (1) and (2)

$\angle 1=\angle 3$ [Corresponding angles for lines $AX$ and $CY$]

$AX||CY$

Hence, proved.