Let the three parallel straight lines AB, CD, EF make equal intercepts KL and LM from the transversal IJ, that is KL = LM.The intercepts made by these three parallel lines on the transversal XY are PQ and QR.Construction:Through Q, a straight line is drawn parallel to IJ to intersect AB and EF at U and V respectively.

Proof:For quadrilateral KLQU,KU∥LQ [∵, AB∥CD] and KL∥UQ [By construction]∴KLQU is a parallelogram.∴KL = UQSimilarly, from quadrilateral LMVQ, we get LM = QV.But it is given that KL = LM.∴UQ = QV.Now, from△UPQ and△QVR, we get∠PUQ = alternate∠QVR [∵AB∥EF, UV is the transversal]∠PQU = vertically opposite∠VQRUQ = QV [Proved before]∴△UPQ≅△VRQ [By A-A-S condition of congruence]∴PQ = QR [Corresponding sides of two congruent triangles]Thus the theorem is proved for three parallel straight line.

Given, KL = LM = MNTo prove, PQ = QR = RS.By drawing a straight line through Q, parallel to IJ, we have proved that PO = QR.Again, a straight line is drawn through R parallel to IJ to intersect CD and GH at Z and W a respectively.As before, if can be proved that QR = RS.∴PQ = QR = RSIn this way, the theorem can be proved for any number of parallel straight lines greater than 3.Remark:From figure 2, we get:KL = LM = MN implies PQ = QR =RS.∴L is the mid point of KM.That is, KM = 2KL.∴KMKL=21or,KMMN=21 [∵KL = MN],∴KM : MN = 2 : 1Similarly from PQ = QR = RS, we get PR : RS = 2: 1So, it can be said:

If three parallel straight lines make two intercepts from a transversal in the ratio 2 :1, then those three parallel straight lines will make two intercepts from transversal in the ratio 2 : 1.