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).


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.



GivenABC and ∆DBC are on the same base BC.
ar(∆ABC) = ar(∆DBC)​
To prove: BC bisects AD
Construction: Draw AL ⊥ BC and DM ⊥ BC.
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 ∆DMOwe have:
ALO = ∠DMO =  90o
∠​AOL = ∠DO​M                  (Vertically opposite angles)
i.e., ∆ ALO ≅ ∆ DMO
∴​ OA = OD
Hence, BC bisects AD.


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