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
Answers
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:
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