Theorem 14.1.3

Theorem: Triangle on the same base and of the same altitude are equal in area.

Given: Triangle ABC and DBC are on the same base $\overline{BC}$ and between the same parallel lines $\overline{BC}$ and $\overline{AD}.$

To Prove:

$\blacktriangle ABC = \blacktriangle DBC$.

Construction:

Draw $\overline{BE}||\overline{CA}$, meeting at $\overline{AD}$ produced at E and also draw $\overline{CF}||\overline{BD}$, meeting at $\overline{AD}$ produced at F.

Proof: