Midpoint Theorem on Right-angled Triangle

Here we will prove that in a right-angled triangle the median drawn to the hypotenuse is half the hypotenuse in length.

Solution:

Given: In ∆PQR, ∠Q = 90°. QD is the median drawn to hypotenuse PR.

$Midpoint Theorem on Right-angled Triangle$

To prove: QS = $\frac{1}{2}$PR.

Construction: Draw ST ∥ QR such that ST cuts PQ at T.

Proof:

Statement	Reason
1. In ∆PQR, PS = $\frac{1}{2}$PR.	1. S is the midpoint of PR.
2. In ∆PQR, (i) S is the midpoint of PR (ii) ST ∥ QR	2. (i) Given. (ii) By construction.
3. Therefore, T is the midpoint of PQ.	3. By converse of the Midpoint Theorem.
4. TS ⊥ PQ.	4. TS ∥ QR and QR ⊥ PQ
5. In ∆PTS and ∆QTS , (i) PT = TQ (ii) TS = TS (iii) ∠PTS = ∠QTS = 90°.	5. (i) From the statement 3. (ii) Common side. (iii) From the statement 4.
6. Therefore, ∆PTS ≅ ∆QTS.	6. By SAS criterion of congruency.
7. PS = QS.	7. CPCTC
8. Therefore, QS = $\frac{1}{2}$PR.	8. Using statement 7 in statement 1.

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Statement	Reason
1. In ∆PQR, PS = \(\frac{1}{2}\)PR.	1. S is the midpoint of PR.
2. In ∆PQR, (i) S is the midpoint of PR (ii) ST ∥ QR	2. (i) Given. (ii) By construction.
3. Therefore, T is the midpoint of PQ.	3. By converse of the Midpoint Theorem.
4. TS ⊥ PQ.	4. TS ∥ QR and QR ⊥ PQ
5. In ∆PTS and ∆QTS , (i) PT = TQ (ii) TS = TS (iii) ∠PTS = ∠QTS = 90°.	5. (i) From the statement 3. (ii) Common side. (iii) From the statement 4.
6. Therefore, ∆PTS ≅ ∆QTS.	6. By SAS criterion of congruency.
7. PS = QS.	7. CPCTC
8. Therefore, QS = \(\frac{1}{2}\)PR.	8. Using statement 7 in statement 1.

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