Step-by-step explanation:
In trapezoid ABCF, AB || CF.
Therefore, parallelograms ABEF & ABCD have same height and lie on the same base AB.
[tex] \therefore Ar(\parallel^{gm} ABEF) = Ar(\parallel^{gm}ABCD)\\... (1)\\\\
\because\: Ar(\parallel^{gm} ABEF) \\=Ar(\triangle AFD) + Ar(\square \: ABED)....(2) \\\\
\&\: Ar(\parallel^{gm}ABCD)\\= Ar(\triangle BCE) +Ar(\square ABED)....(3) [/tex]
From equations (1), (2) & (3), we find:
[tex] Ar(\triangle AFD) + Ar(\square \: ABED) \\ = Ar(\triangle BCE) +Ar(\square ABED)
\\ \\ \purple {\boxed {\therefore Ar(\triangle AFD) = Ar(\triangle BCE)}}
\\ [/tex]
Thus, the two triangular pieces of land are equal in area.
Hence proved.
