Here we will prove that parallelogram on the same base and between the same parallel lines are equal in area.
Given: PQRS and PQMN are two parallelograms on the same base PQ and between same parallel lines PQ and SM.
To prove: ar(parallelogram PQRS) = ar(parallelogram PQMN).
Construction: Produce QP to T.
Proof:
Statement |
Reason |
1. PS = QR. |
1. Opposite sides of the parallelogram PQRS. |
2. PN = QM. |
2. Opposite sides of the parallelogram PQMN. |
3. ∠SPT = ∠RQT. |
3. Opposite sides PS and QR are parallel and TPQ is a transversal. |
4. ∠NPT = ∠MQT. |
4. Opposite sides PN and QM are parallel and TPQ is a transversal. |
5. ∠NPS = ∠MQR. |
5. Subtracting statements 3 and 4. |
6. ∆PSN ≅ ∆RQM |
6. By SAS axiom of congruency. |
7. ar(∆PSN) ≅ ar(∆RQM). |
7. By area axiom for congruent figures. |
8. ar(∆PSN) + ar(quadrilateral PQRN) = ar(∆RQM) + ar(quadrilateral PQRN) |
8. Adding the same area on both sides of the equality in statement 7. |
9. ar(parallelogram PQRS) = ar(parallelogram PQMN). (Proved) |
9. By addition axiom for area. |
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