Mathematics
P and Q are points on opposite sides AD and BC of a parallelogram ABCD such that PQ passes through the point of intersection O of its diagonals AC and BD. Show that PQ is bisected at O.
Answer
Parallelogram ABCD is shown in the figure below:
Considering △OAP and △OCQ we have,
⇒ ∠OAP = ∠OCQ (Alternate angles are equal)
⇒ OA = OC (As diagonals bisect each other)
⇒ ∠AOP = ∠COQ (Vertically opposite angles)
Hence, △OAP ≅ △OCQ by ASA axiom.
OP = OQ (By C.P.C.T.)
Hence, proved that PQ is bisected at O.
Related Questions
In figure (1) given below, ABCD is a parallelogram and X is mid-point of BC. The line AX produced meets DC produced at Q. The parallelogram ABPQ is completed. Prove that
(i) the triangles ABX and QCX are congruent.
(ii) DC = CQ = QP
In figure (2) given below, points P and Q have been taken on opposite sides AB and CD respectively of a parallelogram ABCD such that AP = CQ. Show that AC and PQ bisect each other.
If ABCD is a rectangle in which the diagonal BD bisects ∠B, then show that ABCD is a square.
Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.