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

Proof: 
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.