Question
Triangles ABC and DBC are on the same base BC with vertices A and D on opposite sides of BC such that ar (ΔABC) = ar(ΔDBC). Show that BC bisects AD.
Solution
Given: Triangles ABC and DBC are on the same base BC with vertices A and D on opposite sides of BC such that ar(ΔABC) = ar(ΔDBC).
To Prove: BC bisects AD.
Proof: ar(ΔABC) = ar(ΔDBC) | Given
| Area of a triangle = 
x Base x Corresponding altitude
AM = DN ...(1)
In ΔAMO and ΔDNO,
AM = DN | From (1)
∠AMN = ∠DNO | Each = 90°
∠AOM = ∠DON
| Vertically opposite angles ∴ ΔAMO ⊥ ΔDNO
| AAS congruence rule ∴ AO = DO | CPCT
⇒ BC bisects AD.
