# Midsegment Theorem on Trapezium

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.