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:

p ^{2} = DG ^{2} +DH ^{2} |

q ^{2} = BE ^{2} + BF
^{2} |

r ^{2} = CF ^{2} + CG
^{2} |

s ^{2} = AE ^{2} + AH
^{2} |

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} .