Question

2nd question please as fast as possible

Answer 1

Answer:

Step-by-step explanation:

join HF

since H and F are the mid points of AD and BC respectively

AH=1/2AD   and   BF=1/2BC

now, ABCD is a parallelogram

⇒AD=BC   and AD║BC

⇒1/2AD=1/2BC   and  1/2AD║1/2BC

⇒AH=BF   and   AH║BF

ABFH is a parallelogram

since parallelogram FHAB and ΔFHE are on the same base FH and between same parallels HF and AB

ar(ΔFHE)=1/2ar(parallelogramFHAB)

similarly, we have

ar(ΔFHE)+ar(ΔFGH)=1/2ar(FHAB)+1/2ar(FHDC)

⇒ar(EFGH)=1/2[ar(FHAB)+ar(FHDC)]

⇒ar(EFGH)=ar(ABCD)

hence proved