Theorem: The line segment joining the midpoints of two sides of a triangle is parallel to the third side and equal to half of it.
Given: A triangle PQR in which S and T are the midpoint of PQ and PR respectively.
To prove: ST ∥ QR and ST = \(\frac{1}{2}\)QR
Construction: Draw RU ∥ QP such that RU meets ST produced at U. Join SR.
Proof:
Statement 
Reason 
1. In ∆PST and ∆RUT, (i) PT = TR (ii) ∠PTS = ∠RTU (iii) ∠SPT = ∠TRU 
1. (i) T is the midpoint of PR. (ii) Vertically opposite angles. (iii) Alternate angles. 
2. Therefore, ∆PST ≅ ∆RUT 
2. By AAS criterion of congruency. 
3. Therefore, PS = RU; ST = TU 
3. CPCTC. 
4. But PS = QS 
4. S is the midpoint of PQ. 
5. Therefore, RU = QS and QS ∥ RU. 
5. From statements 3, 4 and construction. 
6. In ∆SQR and ∆RUS, ∠QSR = ∠URS, QS = RU. 
6. From statement 5. 
7. SR = SR. 
7. Common side 
8. ∆SQR ≅ ∆RUS. 
8. SAS criterion of congruency. 
9. QR = SU = 2ST and ∠QRS = ∠RSU 
9. CPCTC and statement 3. 
10. ST = \(\frac{1}{2}\)QR and ST ∥ QR 
10. By statement 9. 
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