In figure 2.30, point T is in the interior of rectangle PQRS, Prove that, \(TS^2 + TQ^2 = TP^2 + TR^2\)
(As shown in the figure, draw seg AB ∥ side SR and A-T-B)
Figure: Rectangle PQRS with P, Q on the top side and S, R on the bottom; A lies on side PS and B lies on side QR.
Answer: Since AB ⊥ PS and AB ⊥ QR, apply Pythagoras theorem in the four right triangles at A and B, and use AP = BQ, AS = BR.
AB ∥ SR ∥ PQ, and PS, QR are perpendicular to SR, so seg AB ⊥ side PS and seg AB ⊥ side QR. Also □ABQP and □ABRS are rectangles, so AP = BQ and AS = BR.
By Pythagoras theorem:
\[TS^2 = TA^2 + AS^2,\quad TQ^2 = TB^2 + BQ^2\] \[TP^2 = TA^2 + AP^2,\quad TR^2 = TB^2 + BR^2\]So
\[TS^2 + TQ^2 = TA^2 + AS^2 + TB^2 + BQ^2\] \[= TA^2 + BR^2 + TB^2 + AP^2\] \[= (TA^2 + AP^2) + (TB^2 + BR^2)\] \[= TP^2 + TR^2\]Hence proved.
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