Question

ABCD is a parallelogram, and AE and CF bisect $\angle A$ and $\angle C$ respectively. Prove that AE || FC.

Answer 1

Given:-

ABCD is a parallelogram, and AE and CF bisect $\angle A$ and $\angle C$ respectively.

To prove:- $AE\parallel CF$

Proof:-

Since in a parallelogram, opposite angles are equal.

$\therefore \angle A=\angle C$

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

$\Rightarrow \angle EAF=\angle FCE⟶\left(1\right)$

Now, $AB\parallel CD$ and CY is the transversal. So,

$\angle FCE=\angle BFC⟶\left(2\right)$

From ${eq}^n\left(1\right)\&\left(2\right)$, we have

$\angle EAF=\angle BFC$

Transversal AB intersects AE and FC at A nad F such that $\angle EAF=\angle BFC$

$\Rightarrow AE\parallel FC$

Hence proved.