Question

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​

Answer 1

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here is your answer..

refer to the aatachment

Answer 2

Given: ABCD is a ||gm

Construction: Join EF

To Prove: AE || CF

Proof:

angle DAB = angle BCD [ Opposite angles of a parallelogram are equal ]

angle DAB / 2 = angle BCD [ Dividing both sides by 2 ]

angle EAF = angle FCE ----------(i)

DC || AB [ since Opposite sides of a ||gm are parallel and equal ]

•°• EC || AF [ E and F are points of AB and CD respectively. ] ----------(ii)

In ∆EAF & ∆FCE ,

angle EAF = angle FCE [ from eq.(i) ]

angle CEF = angle AFE [ EF is transversal and AB || CD ]

EF = EF [ common side ]

•°• by AAS criteria,

∆ECF ≈ ∆FAE

EC = FA [ by C.P.C.T ]

•°• AECF is a parallelogram [ one pair of opposite sides ( EC , AF ) of a ||gm are parallel and equal ]

•°• EC || AF [ opposite sides of a ||gm are parallel ]

Hence, Proved.