in the figure, PQRS is a trepezium in which PQ // SR and PS = QR, show that angle P = angle Q ? ( pls ans urgently and with detailed explanation)
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Given that PQRS is trapezium and PQ ∥SR AND PS∥QR
Let PQ is extended to T. Then, draw a line through R, which is parallel to PS, intersecting PT at point T. It is clear that PTRS is a parallelogram.
(i) PS = RT (Opposite sides of parallelogram PTRS are equal)
However, PS = QR (Given)
Therefore, QR = RT
∠RTQ = ∠RQT (Angle opposite to equal sides are also equal)
Consider the parallel lines PS and RT. PT is the transversal line for them..
∠P + ∠RTQ = 180º ( Sum of angles on the same side are equal to 180 )
∠P + ∠RQT = 180º (Using the relation ∠RTQ = ∠RQT ) ... (1)
However, ∠Q + ∠RQT = 180º (Linear pair angles) ... (2)
From equations (1) and (2), we obtain,
∠P = ∠Q.
Let PQ is extended to T. Then, draw a line through R, which is parallel to PS, intersecting PT at point T. It is clear that PTRS is a parallelogram.
(i) PS = RT (Opposite sides of parallelogram PTRS are equal)
However, PS = QR (Given)
Therefore, QR = RT
∠RTQ = ∠RQT (Angle opposite to equal sides are also equal)
Consider the parallel lines PS and RT. PT is the transversal line for them..
∠P + ∠RTQ = 180º ( Sum of angles on the same side are equal to 180 )
∠P + ∠RQT = 180º (Using the relation ∠RTQ = ∠RQT ) ... (1)
However, ∠Q + ∠RQT = 180º (Linear pair angles) ... (2)
From equations (1) and (2), we obtain,
∠P = ∠Q.
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