Here we will prove that the line segment joining the midpoints of the nonparallel sides of a trapezium is half the sum of the lengths of the parallel sides and is also parallel to them.
Solution:
Given: PQRS is a trapezium in which PQ ∥ RS. U and V are the midpoints of QR and PS respectively.
To prove: (i) UV ∥ RS.
(ii) UV = \(\frac{1}{2}\)(PQ + RS).
Construction: Join QV and produce it to meet RS produced at T.
Proof:
Statement 
Reason 
1. In ∆PQV and ∆STV, (i) PV = VS. (ii) ∠PVQ = ∠TVS. (iii) ∠QPV = ∠VST. 
1. (i) Given. (ii) Vertically opposite angles. (iii) Alternate angles. 
2. Therefore, ∆PQV ≅ ∆STV. 
2. By ASA criterion of congruency. 
3. Therefore, PQ = ST. 
3. CPCTC. 
4. QV = VT. 
4. CPCTC. 
5. In ∆QRT, (i) U is the midpoint of QR. (ii) V is the midpoint of QT. 
5. (i) Given. (ii) From statement 4. 
6. Therefore, UV ∥ RT and UV = \(\frac{1}{2}\)RT. 
6. By the Midpoint Theorem. 
7. Therefore, UV = \(\frac{1}{2}\)(RS+ ST). 
7. From statement 6. 
8. UV = \(\frac{1}{2}\)(RS+ PQ). 
8. Using statement 3 in statement 7. 
9. Therefore, UV ∥ RS and UV = \(\frac{1}{2}\)(PQ+ RS). (Proved) 
9. From statement 6 and 8. 
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