M and N are the mid-points of the sides QR and PQ respectively of a △ PQR, right-angled at Q. Prove that :

(i) PM² + RN² = 5 MN²

(ii) 4 PM² = 4 PQ² + QR²

(iii) 4 RN² = PQ² + 4 QR²

(iv) 4 (PM² + RN²) = 5 PR²

Since, M and N are the mid-points of the sides QR and PQ respectively.

PN = NQ and QM = RM

(i) In △ MNQ,

By pythagoras theorem,

⇒ MN² = NQ² + QM² ……….(1)

In △ PQM,

By pythagoras theorem,

⇒ PM² = PQ² + QM²

⇒ PM² = (PN + NQ)² + QM²

⇒ PM² = PN² + NQ² + 2.PN.NQ + QM²

⇒ PM² = MN² + PN² + 2.PN.NQ [From equation (1)] ………..(2)

In △ RNQ,

By pythagoras theorem,

⇒ RN² = NQ² + RQ²

⇒ RN² = NQ² + (QM + RM)²

⇒ RN² = NQ² + QM² + RM² + 2.QM.RM

⇒ RN² = MN² + RM² + 2.QM.RM [From equation (1)] ………..(3)

Adding equations (2) and (3), we get :

⇒ PM² + RN² = MN² + PN² + 2.PN.NQ + MN² + RM² + 2.QM.RM

⇒ PM² + RN² = 2MN² + PN² + RM² + 2.PN.NQ + 2.QM.RM

Substituting PN = QN and RM = QM in above equation, we get :

⇒ PM² + RN² = 2MN² + QN² + QM² + 2.NQ.NQ + 2.QM.QM

⇒ PM² + RN² = 2MN² + QN² + QM² + 2 QN² + 2 QM²

⇒ PM² + RN² = 2MN² + (QN² + QM²) + 2 (QN² + QM²)

⇒ PM² + RN² = 2MN² + MN² + 2MN² [From equation (1)]

⇒ PM² + RN² = 5MN².

Hence, proved that PM² + RN² = 5MN².

(ii) In △ PQM,

By pythagoras theorem,

⇒ PM² = PQ² + QM²

Multiplying both sides of the above equation by 4, we get :

⇒ 4PM² = 4PQ² + 4QM²

⇒ 4PM² = 4PQ² + $4 \times \Big(\dfrac{1}{2}QR\Big)^2$

⇒ 4PM² = 4PQ² + $4 \times \dfrac{1}{4} \times QR^2$

⇒ 4PM² = 4PQ² + QR².

Hence, proved that 4PM² = 4PQ² + QR².

(iii) In △ RQN,

By pythagoras theorem,

⇒ RN² = NQ² + QR²

Multiplying both sides of the above equation by 4, we get :

⇒ 4RN² = 4NQ² + 4QR²

⇒ 4RN² = 4QR² + $4 \times \Big(\dfrac{1}{2}PQ\Big)^2$

⇒ 4RN² = 4QR² + $4 \times \dfrac{1}{4} \times PQ^2$

⇒ 4RN² = 4QR² + PQ².

Hence, proved that 4RN² = PQ² + 4QR².

(iv) Proved in part (i), we get :

⇒ PM² + RN² = 5MN²

Multiplying both side of the above equation by 4, we get :

⇒ 4(PM² + RN²) = 4 × 5 MN²

⇒ 4(PM² + RN²) = 4 × 5 (NQ² + MQ²)

⇒ 4(PM² + RN²) = 4 × 5 $\times \Big[\Big(\dfrac{1}{2}PQ\Big)^2 + \Big(\dfrac{1}{2}RQ\Big)^2\Big]$

⇒ 4(PM² + RN²) = 4 × 5 $\times \Big[\dfrac{1}{4}PQ^2 + \dfrac{1}{4}RQ^2\Big]$

⇒ 4(PM² + RN²) = 4 × 5 $\times \dfrac{1}{4}$ (PQ² + RQ²)

⇒ 4(PM² + RN²) = 5 (PQ² + RQ²) …..(4)

In right angled triangle PQR,

By pythagoras theorem,

⇒ PR² = PQ² + RQ²

Substituting above value of PR² in equation (4), we get :

⇒ 4(PM² + RN²) = 5 PR².

Hence, proved that 4(PM² + RN²) = 5 PR².