(a) In figure (1) given below, ABCD is a parallelogram and X is the 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
(b) 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.
(a) Given: ABCD is parallelogram and X is mid-point of BC. The line AX produced meets DC produced
at Q and ABPQ is a || gm.
To prove: (i) ∆ABX ≅ ∆QCX
(ii) DC = CQ = QP
Proof:
In ∆ABX and ∆QCX, we have
BX = XC [X is the mid-point of BC]
∠AXB = ∠CXQ [Vertically opposite angles]
∠XCQ = ∠XBA [Alternate angle, since AB || CQ]
So, ABX ≅ ∆QCX by A.S.A axiom of congruence
Now, by C.P.C.T
CQ = AB
But,
AB = DC and AB = QP [As ABCD and ABPQ are || gms]
Hence,
DC = CQ = QP
(b) In || gm ABCD, P and Q are points on AB and CD respectively, PQ and AC intersect each other at O
and AP = CQ
To prove: AC and PQ bisect each other i.e. AO = OC and PO = OQ
Proof:
In ∆AOP and ∆COQ
AP = CQ [Given]
∠AOP = ∠COQ [Vertically opposite angles]
∠OAP = ∠OCP [Alternate angles]
So, ∆AOP ≅ ∆COQ by A.A.S axiom of congruence
Now, by C.P.C.T
OP = OQ and OA = OC
Hence proved.
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