Applying Pythagoras’ theorem we will prove the problem given below.
∆PQR is right-angle at Q. M and N are the midpoints of PQ and QR respectively. Prove that PN2 + RM2 = 5MN2.
Solution:
Given: In ∆PQR, ∠PQR = 90°.
PM = MQ and QN = NR
Therefore, PQ = 2MQ and QR = 2QN
To prove: PN2 + RM2 = 5MN2.
Proof:
Statement |
Reason |
1. ∆PQN, PQ2 + QN2 = PN2 ⟹ (2MQ)2 + QN2 = PN2 ⟹ 4MQ2 + QN2 = PN2 |
1. By Pythagoras’ theorem Given |
2. ∆RQM, MQ2 + QR2 = RM2 ⟹ MQ2 + (2QN)2 = RM2 ⟹ MQ2 + 4QN2 = RM2 |
2. By Pythagoras’ theorem Given |
3. 5MQ2 + 5QN2 = PN2 + RM2 ⟹ 5(MQ2 + QN2) = PN2 + RM2 |
3. Adding statements 1 and 2. |
4. 5MN2 = PN2 + RM2 (Proved) |
4. Applying Pythagoras’ theorem in ∆QMN. |
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