Question

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

Answer 1

Given: ∆ABC and ∆DBC are on the same base BC.
ar(∆ABC) = ar(∆DBC)​
To prove: BC bisects AD
Construction: Draw AL ⊥ BC and DM ⊥ BC.
Proof:
Since ∆ABC and ∆DBC are on the same base BC and they have equal areas, their altitudes must be equal.
i.e., AL = DM
Let AD and BC intersect at O.
Now, in ∆ALO and ∆DMO, we have:
AL = DM
∠ALO = ∠DMO = 90^o
∠​AOL = ∠DO​M (Vertically opposite angles)
i.e., ∆ ALO ≅ ∆ DMO
​​∴​ OA = OD
Hence, BC bisects AD.