Here we will prove that in a square, the diagonals are equal in length and they meet at right angles.
Given: PQRS is a square in which PQ = QR = RS = SP, and ∠QPS = ∠PQR = ∠QRS = ∠RSP = 90°.
To prove: PR = QS and PR ⊥ QS
Proof:
Statement |
Reason |
1. In ∆SPQ and ∆RQP, (i) SP = QR |
(i) Given |
(ii) PQ = PQ |
(ii) Common side |
(iii) ∠SPQ = ∠PQR |
(iii) Given |
(iv) ∆SPQ ≅ ∆RQP Therefore, QS = PR (Proved) |
(iv) By SAS criterion of congruency. CPCTC. |
2. (v) ∠PQS = ∠PSQ |
(v) In ∆PQS, PQ = PS |
(vi) ∠PQS + ∠PSQ = 90°. |
(vi) In ∆QPS, ∠QPS = 90° and sum of three angles of a triangle is 180°. |
(vii) ∠PQS = \(\frac{90°}{2}\) = 45° |
(vii) By statements (v) and (vi). |
(viii) ∠QPR = 45° |
(viii) Similarly as (vi) and (vii) for the ∆PQR. |
(ix) ∠POQ = 180° - (PQO + ∠QPO) = 180° - (45° + 45°) = 180° - 90° = 90° Therefore, OP ⊥ OQ Therefore, ∠POQ = 90° Therefore, PR ⊥ QS (Proved) |
(ix) By statements (vii), (viii) and the sum of the angles of ∆POQ is 180°. |
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