Question

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

Answer 1

Answer: AE || CF

Step-by-step explanation: Given,

ABCD is a ||gm , linesegment AE and CF bisects <A and <C .

To proof,

AE || CF

Proof,

(1) <A = <C

=> 1/2 <A = 1/2 <C

=> <1 = <2

(2) AB||CD and CY is the transversal

So, <2=<3.

From equation (1) and (2) we have ,

=> <1 = <2

Transversal AB intersect AE and CF at A and F such that ,

=> <1 = <3

= AE||CF

Hence , proved !!

Answer 2

Given :-

ABCD is a parallelogram

AE bisect ∠A

CF bisect ∠C

To Proof :-

AE || CF

Solution :-

Since it is given ABCD is a parallelogram

Therefore, ∠A = ∠C [ Opposite angle of parallelogram are equal ]

-> 1/2 ∠A = 1/2 ∠B

∠1 = ∠2 ---(1)

AE is bisector of ∠A and FC is bisector of ∠C

AB || DC [Opposite side of parallelogram are parallel ]

∠2 = ∠3 ---(2) [Alternate interior angle ]

From equation (1)& (2) We get,

∠1 = ∠3

Since corresponding angle are equal therefore, AE || FC.