`X Y` isa line parallel to side `B C` of ` A B CdotB E A` and `C F A B` meet `X Y` in `E` and `F` respectively.Show that `a r( A B E)=a r( A C F)dot`

Explanation

Given: `/_\ABC`, `BE|\|AC` and `CF|\|AB`
To prove: `ar(ABE)=ar(ACF)`
Proof: since `YE|\|BC` and `BE||CY`
`BCYE` is a parallelogram.
Similarly, `BCFX` is a parallelogram.
Parallelograms `BCFX` and `BCYE` are on the same base `BC` and between same parallels `BC` and `EF`,
`:.ar(BCYE)=ar(BCFX)` …(i)
`ar(ABE)=1/2ar(BCYE)` …(ii) (on same base and between same parallels)
Similarly, `ar(ACF)=1/2ar(BCFX)` …(iii)
From (i), (ii) and (iii), we get:
`ar(ABE)=ar(ACF)`
Hence proved.