Question

prove that Δ's on the same base and between the same parallels are equal in area ​

Answer 1

Let ∆ABC and ∆ABD be on the same base AB and between the same parallel AB and CD. It is require to prove that ∆ABC = ∆ABD.

Construction: A parallelogram ABPQ is constructed with AB as base and lying between the same parallels AB and CD.

Triangles on Same Base and between Same Parallels

Proof: Since ∆ABC and parallelogram ABPQ are on the same base AB and between the same parallels AB and Q,

Therefore, ∆ABC = ½(Parallelogram ABPQ)

Similarly, ∆ABD = ½(Parallelogram ABPQ)

Therefore, ∆ABC = ∆ABD.

Answer 2

Given triangles ABC and DBC are on the same base BC. Consider, Δ’s ABC and DBC ∠A = ∠D = 90° (Given) ∠AEB = ∠DEC (Vertically opposite angles are equal) Hence ΔABC ~ ΔDBC (AA similarity theorem) ∴ AE × CE = BE × DE