Traingles ABC & DBC are on the same base BC with vertices A & D on opposite sides of BC such that ar ABC = ar DBC. Show that BC bisects AD.
Given: Δ ABC and Δ BCD, are on the same base BC. A and D are on opposite sides of BC and area ( Δ ABC) = area ( ΔBCD)
Construction: Draw perpendicular AM and DN on BC from A and D respectively.
To prove: AO = OD
area ( ΔABC) = area ( ΔBCD) (Given)
⇒ AM = DN
let AD and BC intersect at O.
In Δ AMO and Δ DNO,
∠AMO = ∠DNO = 90 o
∠AOM = ∠DON [vertically opposite angles]
AM = DN [proved above]
⇒ Δ AMO Δ DNO [AAS congruence criterion]
∴ AO = OD [Corresponding parts of congruent triangles]
Hence, BC bisects AD.