PQRS is a trapezium in which PQ ∥ RS. T is the midpoint of QR. TU is drawn parallel to PQ which meets PS at U. Prove that 2TU = PQ + RS.
Given: PQRS is a trapezium in which PQ ∥ RS. T is the midpoint of QR. TU ∥ PQ and TU meets PS at U.
To prove: 2TU = PQ + RS.
Construction: Join QS. QS and TU intersect at M.
Proof:
Statement 
Reason 
1. PQ ∥ RS and TU ∥ PQ. 
1. Given. 
2. RS ∥ TU. 
2. From statement 1. 
3. In ∆QRS, T is the midpoint of QR and TM ∥ RS ⟹ M is the midpoint of QS. 
3. By the converse of the Midpoint Theorem. 
4. In ∆PSQ, M is the midpoint of QS and MU ∥ PQ. ⟹ U is the midpoint of PS. 
4. By the converse of the Midpoint Theorem. 
5. In ∆QRS, the line segment TM joining the midpoints of sides QR and QS. Therefore, TM = \(\frac{1}{2}\)RS. 
5. By the Midpoint Theorem. 
6. In ∆PQS, the line segment MU joins the midpoints of the sides QS and PS. Therefore, MU = \(\frac{1}{2}\)PQ. 
6. By the Midpoint Theorem. 
7. TM + MU = \(\frac{1}{2}\)RS + \(\frac{1}{2}\)PQ. 
7. From statements 5 and 6. 
8. TU = \(\frac{1}{2}\)(RS + PQ). 
8. TM + MU = TU. 
9. 2TU = RS + PQ. (Proved) 
9. From statement 8. 
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