Question

Triangles ABC and DBC are on the same base BC; with A, D on opposite sides of the line BC, such that ar(∆ABC) = ar(∆DBC). Show that BC bisects AD.

Answer 1

Construction: Draw AL ⊥ BC and DM ⊥ BC.

Proof : ar(∆ABC) = ar(∆DBC) [Given]

⇒ BC x AL /2 = BC x DM/2

⇒ AL = DM ....(i)

Now in ∆s OAL and OMD

AL = DM [From (i)]

⇒ ∠ALO = ∠DMO [Each = 900 ]

⇒ ∠AOL = ∠MOD [Vert. opp. ∠s]

⇒ ∠OAL = ∠ODM [Third angles of the triangles]

∴ ∆OAL ≅ ∆OMD [By ASA]

∴ OA = OD [By cpctc]

i.e., BC bisects AD