prove that two triangles having the same base (or equal bases) and equal areas lie between the same parallels.
Given: Δ ABC and Δ PBC lie on the same base BC and having equal areas.
To prove: ΔABC and ΔPBC lies between the same parallels l and m.
Construction: Draw AD and PQ, the altitude drawn from A and P on BC respectively.
Given, ar (ΔABC) = ar(ΔPBC)
We know that area of a triangle
Therefore, Area (ΔABC)
and Area (ΔPBC)
On equating (1) and (2), we get
⇒ The height of perpendiculars drawn between two lines l and m are equal which is possible only in one case when the lines l and m are running parallel to each other.
Hence, the triangles ABC and PBC drawn on the same base BC and having equal areas lying between the same parallels l and m.