Diagonals of a Square are Equal in Length and they Meet at Right Angles

            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°.