Question

plzz solve it quickly

Answer 1

Solution :

Solution : Given,

Solution : Given,XY || BC, BE || AC and CF || AB

Solution : Given,XY || BC, BE || AC and CF || ABTo show,

Solution : Given,XY || BC, BE || AC and CF || ABTo show,ar(ΔABE) = ar(ΔAC)

Solution : Given,XY || BC, BE || AC and CF || ABTo show,ar(ΔABE) = ar(ΔAC)Proof:

Solution : Given,XY || BC, BE || AC and CF || ABTo show,ar(ΔABE) = ar(ΔAC)Proof:EY || BC (XY || BC) --- (i)

Solution : Given,XY || BC, BE || AC and CF || ABTo show,ar(ΔABE) = ar(ΔAC)Proof:EY || BC (XY || BC) --- (i)BE∥ CY (BE || AC) --- (ii)

Solution : Given,XY || BC, BE || AC and CF || ABTo show,ar(ΔABE) = ar(ΔAC)Proof:EY || BC (XY || BC) --- (i)BE∥ CY (BE || AC) --- (ii)BEYC is a parallelogram

Solution : Given,XY || BC, BE || AC and CF || ABTo show,ar(ΔABE) = ar(ΔAC)Proof:EY || BC (XY || BC) --- (i)BE∥ CY (BE || AC) --- (ii)BEYC is a parallelogramBXFC is a parallelogram.

Solution : Given,XY || BC, BE || AC and CF || ABTo show,ar(ΔABE) = ar(ΔAC)Proof:EY || BC (XY || BC) --- (i)BE∥ CY (BE || AC) --- (ii)BEYC is a parallelogramBXFC is a parallelogram.Parallelograms on the same base BC and between the same parallels EF and BC.

Solution : Given,XY || BC, BE || AC and CF || ABTo show,ar(ΔABE) = ar(ΔAC)Proof:EY || BC (XY || BC) --- (i)BE∥ CY (BE || AC) --- (ii)BEYC is a parallelogramBXFC is a parallelogram.Parallelograms on the same base BC and between the same parallels EF and BC.⇒ ar(BEYC) = ar(BXFC) (Parallelograms on the same base BC and between the same parallels EF and BC)

Solution : Given,XY || BC, BE || AC and CF || ABTo show,ar(ΔABE) = ar(ΔAC)Proof:EY || BC (XY || BC) --- (i)BE∥ CY (BE || AC) --- (ii)BEYC is a parallelogramBXFC is a parallelogram.Parallelograms on the same base BC and between the same parallels EF and BC.⇒ ar(BEYC) = ar(BXFC) (Parallelograms on the same base BC and between the same parallels EF and BC) Also,

Solution : Given,XY || BC, BE || AC and CF || ABTo show,ar(ΔABE) = ar(ΔAC)Proof:EY || BC (XY || BC) --- (i)BE∥ CY (BE || AC) --- (ii)BEYC is a parallelogramBXFC is a parallelogram.Parallelograms on the same base BC and between the same parallels EF and BC.⇒ ar(BEYC) = ar(BXFC) (Parallelograms on the same base BC and between the same parallels EF and BC) Also,△AEB and parallelogram BEYC are on the same base BE and between the same parallels BE and AC.

Solution : Given,XY || BC, BE || AC and CF || ABTo show,ar(ΔABE) = ar(ΔAC)Proof:EY || BC (XY || BC) --- (i)BE∥ CY (BE || AC) --- (ii)BEYC is a parallelogramBXFC is a parallelogram.Parallelograms on the same base BC and between the same parallels EF and BC.⇒ ar(BEYC) = ar(BXFC) (Parallelograms on the same base BC and between the same parallels EF and BC) Also,△AEB and parallelogram BEYC are on the same base BE and between the same parallels BE and AC.=) ar(△AEB) = 1/2ar(BEYC) --- (iv)

Solution : Given,XY || BC, BE || AC and CF || ABTo show,ar(ΔABE) = ar(ΔAC)Proof:EY || BC (XY || BC) --- (i)BE∥ CY (BE || AC) --- (ii)BEYC is a parallelogramBXFC is a parallelogram.Parallelograms on the same base BC and between the same parallels EF and BC.⇒ ar(BEYC) = ar(BXFC) (Parallelograms on the same base BC and between the same parallels EF and BC) Also,△AEB and parallelogram BEYC are on the same base BE and between the same parallels BE and AC.=) ar(△AEB) = 1/2ar(BEYC) --- (iv)△ACF and parallelogram BXFC

Solution : Given,XY || BC, BE || AC and CF || ABTo show,ar(ΔABE) = ar(ΔAC)Proof:EY || BC (XY || BC) --- (i)BE∥ CY (BE || AC) --- (ii)BEYC is a parallelogramBXFC is a parallelogram.Parallelograms on the same base BC and between the same parallels EF and BC.⇒ ar(BEYC) = ar(BXFC) (Parallelograms on the same base BC and between the same parallels EF and BC) Also,△AEB and parallelogram BEYC are on the same base BE and between the same parallels BE and AC.=) ar(△AEB) = 1/2ar(BEYC) --- (iv)△ACF and parallelogram BXFC⇒ ar(△ ACF) = 1/2ar(BXFC) --- (v)

Solution : Given,XY || BC, BE || AC and CF || ABTo show,ar(ΔABE) = ar(ΔAC)Proof:EY || BC (XY || BC) --- (i)BE∥ CY (BE || AC) --- (ii)BEYC is a parallelogramBXFC is a parallelogram.Parallelograms on the same base BC and between the same parallels EF and BC.⇒ ar(BEYC) = ar(BXFC) (Parallelograms on the same base BC and between the same parallels EF and BC) Also,△AEB and parallelogram BEYC are on the same base BE and between the same parallels BE and AC.=) ar(△AEB) = 1/2ar(BEYC) --- (iv)△ACF and parallelogram BXFC⇒ ar(△ ACF) = 1/2ar(BXFC) --- (v)From (iii), (iv) and (v),

Solution : Given,XY || BC, BE || AC and CF || ABTo show,ar(ΔABE) = ar(ΔAC)Proof:EY || BC (XY || BC) --- (i)BE∥ CY (BE || AC) --- (ii)BEYC is a parallelogramBXFC is a parallelogram.Parallelograms on the same base BC and between the same parallels EF and BC.⇒ ar(BEYC) = ar(BXFC) (Parallelograms on the same base BC and between the same parallels EF and BC) Also,△AEB and parallelogram BEYC are on the same base BE and between the same parallels BE and AC.=) ar(△AEB) = 1/2ar(BEYC) --- (iv)△ACF and parallelogram BXFC⇒ ar(△ ACF) = 1/2ar(BXFC) --- (v)From (iii), (iv) and (v),ar(△AEB) = ar(△ACF).

Answer 2

hey dear your answer is here.
see attachment