In figure, ABC and ABD are two triangles on the base AB. If line segment CD is bisected by AB at O, show that ar(ΔABC) = ar(ΔABD).
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Draw two perpendiculars CP and DQ on AB.
Now,
ar(ΔABC) = 1/2×AB×CP ⋅⋅⋅⋅⋅⋅⋅(1)
ar(ΔABD) = 1/2×AB×DQ ⋅⋅⋅⋅⋅⋅⋅(2)
To prove the result, ar(ΔABC) = ar(ΔABD), we have to show that CP = DQ.
In right angled triangles, ΔCPO and ΔDQO
∠CPO = ∠DQO = 90o
CO = OD (Given)
∠COP = ∠DOQ (Vertically opposite angles)
By AAS condition: ΔCP0 ≅ ΔDQO
So, CP = DQ …………..(3)
(By CPCT)
From equations (1), (2) and (3), we have
ar(ΔABC) = ar(ΔABD)
Hence proved.
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