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 $\left(△ABC\right)=ar\left(△DBC\right)$ 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 .