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Clearly, PQRS is a parallelogram having one of its interior angles as 90°., Hence, PQRS is a rectangle., , Q3. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC,, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus., , Difficulty Level:, Medium, , What is known/given?, ABCD is a rectangle and P, Q, R and S are the mid-points of the sides AB, BC, CD and, DA respectively., , What is unknown?, How we can show that quadrilateral PQRS is a rhombus., , Reasoning:, , Ina triangle, the line segment joining the mid-points of any two sides of the triangle is, parallel to the third side and is half of it. Also if one pair of opposite sides of quadrilateral, is parallel and equal to each other then it is a parallelogram. By showing all four sides, equal in parallelogram we can say it is rhombus., , , , , , , , Solution:, D, R c, s| | Q, A Pp B, , Let us join AC and BD., In AABC,, P and Q are the mid-points of AB and BC respectively., PQ || AC and PQ = ; AC (Mid-point theorem) ... (1), Similarly, in AADC,, SR || AC and SR = pac (Mid-point theorem) ... (2), Clearly, PQ |] SR and PQ = SR, Since in quadrilateral PQRS, one pair of opposite sides is equal and parallel to each other,, itis a parallelogram., =-PS || QR and PS = QR (Opposite sides of parallelogram) ... (3), In ABCD, Q and R are the mid-points of side BC and CD respectively., ©.QR |! BD and QR = 5BD (Mid-point theorem) ... (4), However, the diagonals of a rectangle are equal., “AC = BD... (5), By using Equations (1), (2), 3), (4), and (5), we obtain, , PQ = QR = SR =PS, ‘Therefore, PQRS is a rhombus., , Q4. ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid point of AD. A line is drawn through E parallel to AB intersecting BC at F, (see the given figure). Show that F is the mid-point of BC., , D Cc, , 7 =