A 3D ANALOGUE OF PYTHAGORAS'S THEOREM

PRESENTED BY

PETER WAKEFIELD SAULT

Copyright © Peter Wakefield Sault 1976-2011

*The proof presented here was devised by the author in 1976.*

The reader is surely already familiar with Pythagoras's Theorem, which will be restated here without proof - that, in any right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides. This will be assumed to be true throughout this work.

The application of Pythagoras's Theorem in Three Dimensions is well-known and involves the relationship between the perpendicular edges of a rectangular block and the solid diagonal of that same block.

In Figure 2 above, (h) is the solid diagonal of a rectangular block.

i. |
h² = a² + d² | (Pythagoras) | ||||

ii. |
but | d² = b² + c² | (Pythagoras) | |||

iii. |
\ | h² = a² + b² + c² | Q.E.D. |

What we are about to do is to assume a 3D analogue of a right-angled
triangle and to show that a statement similar to Pythagoras's Theorem can be
made about it. The 3D analogue of a right-angled triangle is an irregular
tetrahedron with a cubic vertex. By ‘*cubic vertex*’ is meant that all
three faces adjacent to the vertex are right-angled triangles and that all
three right angles touch the vertex. This can be imagined as any single
corner cut from a rectangular block so as to produce an irregular
tetrahedron.

To prove that:-

**In any tetrahedron with a cubic vertex the numerical square of the area of the
face opposite the cubic vertex is equal to the sum of the numerical squares of
the areas of the other three faces.**

Let:-

A = Area of Triangle OXZ

B = Area of Triangle OXY

C = Area of Triangle OYZ

D = Area of Triangle XYZ

To prove that D² = A² + B² + C²

I. |
D = ½ah | (given) | ||||

II. |
Now | PX = Ö(b² - h²) | (Pythagoras) | |||

III. |
and | PZ = Ö(c² - h²) | (Pythagoras) | |||

IV. |
and | a = PX + PZ | ||||

V. |
\ | a = Ö(b² - h²) + Ö(c² - h²) | ||||

VI. |
\ | a² = b² + c² - 2h² + 2Ö(b²c² - b²h² - c²h² + h²) | ||||

VII. |
\ | a² - b² - c² + 2h² = 2Ö(b²c² - b²h² - c²h² + h²) | ||||

VIIIa. |
\ | (a² - b² - c² + 2h²)² = 4(b²c² - b²h² - c²h² + h²) | ||||

VIIIb. |
i.e. | (a² - b² - c²)² + (a² - b² - c²)4h² + 4h² = 4b²c² - 4b²h² - 4c²h² + 4h² | ||||

IX. |
\ | (a² - b² - c²)² + 4a²h² = 4b²c² | ||||

X. |
\ | 4a²h² = 4b²c² - (a² - b² - c²)² | ||||

XI. |
but | D = ½ah | (I.) | |||

XII. |
\ | 16D² = 4b²c² - (a² - b² - c²)² | ||||

XIII. |
but | a² = x² + z² | (Pythagoras) | |||

XIV. |
and | b² = x² + y² | (Pythagoras) | |||

XV. |
and | c² = y² + z² | (Pythagoras) | |||

XVI. |
\ | 16D² = 4(x² + y²)(y² + z²) - 4y^{4} |
||||

XVII. |
\ | 4D² = x²y² + x²z² + y²z² | ||||

XVIII. |
Now | A = ½xz | (given) | |||

XIX. |
and | B = ½xy | (given) | |||

XX. |
and | C = ½yz | (given) | |||

XXI. |
\ | A + B + C = ½(xy + xz + yz) | ||||

XXII. |
\ | A² + B² + C² = ¼(x²y² + x²z² + y²z²) | ||||

XXIII. |
\ | 4(A² + B² + C²) = x²y² + x²z² + y²z² | ||||

XXIV. |
but | 4D² = x²y² + x²z² + y²z² | (XVII.) | |||

XXV. |
\ | 4D² = 4(A² + B² + C²) | ||||

XXVI. |
\ | D² = A² + B² + C² | Q.E.D. |

There is another proof which, unlike that given above, is neither so abstruse nor of my own devising (1976). For this proof a tetrahedron OXYZ with cubic vertex at O is cut by a plane through edge OY such that edge XZ is normal to it. The plane intersects edge XZ at P and angles OPX, OPZ, YPX and YPZ are all right angles.

Let:-

A = Area of Triangle OXZ

B = Area of Triangle OXY

C = Area of Triangle OYZ

D = Area of Triangle XYZ

To prove that A² + B² + C² = D²

I. |
Now | A = ½ag | (given) | |||

II. |
and | B = ½xy | (given) | |||

III. |
and | C = ½yz | (given) | |||

IV. |
\ | A² + B² + C² = ¼a²g² + ¼y²(x² + z²) | ||||

V. |
but | a² = x² + z² | (Pythagoras) | |||

VI. |
\ | A² + B² + C² = ¼a²(g² + y²) | ||||

VII. |
but | h² = g² + y² | (Pythagoras) | |||

VIII. |
\ | A² + B² + C² = (½ah)² | ||||

IX. |
but | D = ½ah | (given) | |||

X. |
\ | A² + B² + C² = D² | Q.E.D. |

Copyright © Peter Wakefield Sault 1976-2011