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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