Prove that triangle on the same …
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Gaurav Seth 4 years, 3 months ago
Let ΔABP and a parallelogram ABCD be on the same base AB and between the same parallels AB and PC.
To Prove : ar( ΔPAB ) = (1/2)ar( ABCD)
(here, ar = area; written in short)
Draw BQ ||AP to obtain another parallelogram.ABQP and ABCD are on the same base AB and between the same parallels AB and PC.
There fore, ar(ABQP) = ar(ABCD)
But ΔPAB ≅ ΔBQP( Diagonals PB divides parallelogram ABQP into two congruent triangles.
So ar (PAB) = ar(BQP) -----------(2)
∴ ar (PAB) = (1/2)ar(ABQP) -----------------(3) [ from (2)]
This gives ar (PAB) = (1/2)ar(ABCD) [ from (1) and (3)]
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