Question

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

Answer 1

Solution:

Since opposite angles are equal in a parallelogram . Therefore , in parallelogram $ABCD$ , we have

$\angle A=\angle C$

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

$\angle 1=\angle 2---\left(1\right)$

[ $AX$ and $CY$ are bisectors of $\angle A$ and $\angle C$ respectively]

Now, $AB\parallel DC$ and the transversal $CY$ intersects them.

$\angle 2$ and $\angle 3$ $---\left(2\right)$ [ alternate interior angles are equal ]

From (i) and (ii) , we have

$\angle 1$ and $\angle 3$

Thus , transversal $AB$ intersects $AX$ and $YC$ at $A$ and $Y$ such that $\angle 1=\angle 3$ i.e. corresponding angles are equal .

$AX\parallel CY$.

Since $AX\parallel CY$ We get $AE=CF$