In Figure, `A B C D E` is a pentagon. A line through `B` parallel to `A C` meets `D C` produced at `Fdot` Show that `a r( A C B)=a r( A C F)` (ii) `a r( A E D F)=a r(A B C D E)`

Explanation

`(i)"A pentagon ABCDE, BF is parallel to AC"`

`"To Prove: Ar(ACB)=Ar(ACF)"`

`Ar(AEDF)=Ar(ABCDE)`

`/_\ACB and /_\ACF" lies on same base AC and are between same parallel AC and BF"`

`∴ Ar(/_\ACB)=Ar(/_\ACF)->"eq(1)"`

`(ii)" add Ar(AEDC) both sides"`

`Ar(/_\ACB)+Ar(AEDC)=Ar(/_\ACF)+Ar(AEDC)`

`Ar(ABCDE)=Ar(AEDF)`