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:
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Statement |
Reason |
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1. In ∆PQV and ∆STV, (i) PV = VS. (ii) ∠PVQ = ∠TVS. (iii) ∠QPV = ∠VST. |
1. (i) Given. (ii) Vertically opposite angles. (iii) Alternate angles. |
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2. Therefore, ∆PQV ≅ ∆STV. |
2. By ASA criterion of congruency. |
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3. Therefore, PQ = ST. |
3. CPCTC. |
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4. QV = VT. |
4. CPCTC. |
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5. In ∆QRT, (i) U is the midpoint of QR. (ii) V is the midpoint of QT. |
5. (i) Given. (ii) From statement 4. |
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6. Therefore, UV ∥ RT and UV = \(\frac{1}{2}\)RT. |
6. By the Midpoint Theorem. |
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7. Therefore, UV = \(\frac{1}{2}\)(RS+ ST). |
7. From statement 6. |
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8. UV = \(\frac{1}{2}\)(RS+ PQ). |
8. Using statement 3 in statement 7. |
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9. Therefore, UV ∥ RS and UV = \(\frac{1}{2}\)(PQ+ RS). (Proved) |
9. From statement 6 and 8. |
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