Question

PQ and RS are two equal and parallel line-segments. Any point M not lying on PQ or RS is joined to Q and S and lines through P parallel to QM and through R parallel to SM meet at N. Prove that line segments MN and PQ are equal and parallel to each other.

Answer 1

We draw the figure as per the given conditions (Fig.).

It is given that PQ = RS and PQ || RS. Therefore, PQSR is a parallelogram. 

So, PR = QS and PR || QS (1) 

Now, PR || QS 

Therefore, ∠RPQ + ∠PQS = 180º (Interior angles on the same side of the transversal) 

i.e., ∠RPQ + ∠PQM + ∠MQS = 180º (2) 

Also, PN || QM (By construction) 

Therefore, ∠NPQ + ∠PQM = 180º 

i.e., ∠NPR + ∠RPQ + ∠PQM = 180º (3) 

So, ∠NPR = ∠MQS [From (2) and (3)] (4) 

Similarly, ∠NRP = ∠MSQ (5) 

Therefore, ∆PNR ≅ ∆QMS [ASA, using (1), (4) and (5)] 

So, PN = QM and NR = MS (CPCT) 

As, PN = QM and 

PN || QM, we have PQMN is a parallelogram 

So, MN = PQ and NM || PQ