Question

XY is a line parallel to side BC of a triangle ABC .If BE Parallel to AC and CF Parallel and AB new XY at E and F respectively , show that ar(ABE) = ar(ACF)

Answer 1

EY || BC (XY || BC) — (i)
also,
BE∥ CY (BE || AC) — (ii)


From (i) and (ii),
BEYC is a parallelogram.

(Both the pairs of opposite sides are parallel.)


Similarly,We can prove that BXFC is a parallelogram.


Parallelograms on the same base BC and between the same parallels EF and BC.


 ar(BEYC) = ar(BXFC)....(iii)

 

(Parallelograms on the same base BC and between the same parallels EF and BC)

 
Also,
△ABE and parallelogram BEYC are on the same base BE and between the same parallels BE and AC.


ar(△ABE) = 1/2ar(BEYC) — (iv)


Similarly,
△ACF & ||gm  BXFC on the same base CF and between the same parallels CF and AB.


ar(△ ACF) = ½ ar(BXFC) — (v)


From (iii), (iv) and (v),
ar(△ABE) = ar(△ACF)

Answer 2

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See the attached files

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