Page 1 - IX Lesson Note -2 triangles on same base-ch 9(Area of Parallelograms and Triangles

SAI International School
CLASS -IX

Mathematics
Chapter-9: Area of Parallelograms and Triangles
Lesson Notes-1

Sub Topic: Theorem 2

Given: Two triangles ABC and PBC on the same base BC and between the same
parallel lines BC and AP.

To prove: ar( ABC) = ar( PBC)

Construction: Through B, draw BD||CA intersecting PA produced in D and through
C, draw CQ||BP, intersecting line AP in Q.

Proof: We have,
BD||CA [by construction]
and BC||DA [given]
Therefore, BCAD is a parallelogram.
Similarly, BCQP is a parallelogram.

Now, parallelogram BCQP and parallelogram BCAD are on the same base BC and
between the same parallels.

Therefore, area(parallelogram BCQP) = area(parallelogram BCAD) (1)

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