Areas of Parallelograms and Triangles

Question

In figure, PS = QR and ∠SPQ = ∠RQP. Prove that PR = QS and ∠QPR = ∠PQS.


Answer

In ∆QPR and ∆PQS,
QR = PS    | Given
∠RQP = ∠SPQ    | Given
PQ = PQ    | Common
∴ ∆QPR ≅ ∆PQS    | SAS Axiom
∴ PR = QS    | C.P.C.T.
and    ∠QPR = ∠PQS.    | C.P.C.T.

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