Question:
Show that a diagonal divides a parallelogram into two triangles of equal area.
Solution:
Let ABCD be a parallelogram and BD be its diagonal.
To prove: ar(∆ABD) = ar(∆CDB)
Proof:
In ∆ABD and ∆CDB, we have:
AB = CD [Opposite sides of a parallelogram]
AD = CB [Opposite sides of a parallelogram]
BD = DB [Common]
i.e., ∆ABD ≅ ∆CDB [ SSS criteria]
∴ ar(∆ABD) = ar(∆CDB)
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