Question

In the figure, ABCD is a parallelogram. Point A, B,
E and F are collinear and AB = EF. Prove that
DEPC is a parallelogram. Step by step explanation please​

Answer 1

Step-by-step explanation:

as ABCD is parallelogram so, AB=DC

but, AB=EF so, EF=DC.

take triangle DCO and BOE,here I have taken the intersection point of line CB AND DE as o.

angle DOC = BOE ( V.O.A)

ANGLE (CDO )=ANGLE( BEO) (A.I.A)

hence,the last angle will also be equal so, triangle DCO CONGURENT TO triangle BOE.

take triangle DAD AND CBF,

ANGLE (A) =ANGLE (CBF) (CORROSPONDING ANGLES).

AB=CB ( PARALLELOGRAM property).

AB+BE =BE+EF

SO, AE =BF

hence, triangle DAD AND CBF, are congurent by SAS.

hence, DE=CF.

HENCE,proved

Answer 2

Answer:

As ABCD is parallelogram so, AB=DC

but, AB=EF so, EF=DC.

take triangle DCO and BOE,here I have taken the intersection point of line CB AND DE as o.

angle DOC = BOE ( V.O.A)

ANGLE (CDO )=ANGLE( BEO) (A.I.A)

hence,the last angle will also be equal so, triangle DCO CONGURENT TO triangle BOE.

take triangle DAD AND CBF,

ANGLE (A) =ANGLE (CBF) (CORROSPONDING ANGLES).

AB=CB ( PARALLELOGRAM property).

AB+BE =BE+EF

SO, AE =BF

hence, triangle DAD AND CBF, are congurent by SAS.

hence, DE=CF.

HENCE,proved