Mathematics
In the adjoining figure, ABCD is a parallelogram and X is the mid-point of BC. The line AX produced meets DC produced at Q. The parallelogram AQPB is completed. Prove that :
(i) ΔABX ≅ ΔQCX.
(ii) DC = CQ = QP.
Rectilinear Figures
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Answer
(i) Considering △ABX and △QCX we have,
⇒ ∠XBA = ∠XCQ (Alternate angles are equal)
⇒ XB = XC (As X is mid-point of BC)
⇒ ∠AXB = ∠CXQ (Vertically opposite angles are equal)
Hence, △ABX ≅ △QCX by ASA axiom.
(ii) Since, △ABX ≅ △QCX
∴ AB = CQ (By C.P.C.T.C.) …….(1)
AB = CD and AB = QP (Opposite sides of parallelogram are equal) ……..(2)
From (i) and (ii) we get,
⇒ AB = DC = CQ = QP
⇒ DC = CQ = QP
Hence, proved that DC = CQ = QP.
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