A proof of this result came from students from the Key Stage 3 Maths Club at Strabane Grammar School, Northern Ireland and the following one from Joel, ACS (Barker) Singapore:

(a) In the diagram given (in 2 dimensions) p ² + q ² = r ² + s ²

Proof

Draw 2 lines through V parallel to the edges of the rectangle, dividing the figure into four pairs of right-angled triangles. Label the point on AB as E, on BC as F, on CD as G and on DA as H. By Pythagoras theorem:

Since DG=AE, CG=BE, AH=BF and CF=DH,

p ² + q ² = AE ² + CF ² + CG ² + AH ² = r ² + s ²

(b) If the diagram represents a pyramid on a rectangular base where p, q, r and s are the lengths of the sloping edges then the result p ² + q ² = r ² + s ² still holds true.

Proof

Let V ₁ be the foot of the perpendicular from V to the base ABCD of the pyramid and let h be the height of the pyramid so that VV ₁ = h and let V ₁ A = s ₁ , V ₁ B = q ₁ , V ₁ C = r ₁ , and V ₁ D= p ₁ .

By Pythagoras theorem we have: p ₁ ² + h ² = p ² , q ₁ ² + h ² = q ² , r ₁ ² + h ² = r ² and s ₁ ² + h ² = s ² .

Using the result already proved in 2dimensions, that is

p ₁ ² + q ₁ ² = r ₁ ² + s ₁ ² ,

we get p ₁ ² + q ₁ ² + 2h ² = r ₁ ² + s ₁ ² + 2h ²

so p ² + q ² = r ² + s ² .