3D ANALOGUE of PYTHAGORAS' THEOREM

Document date: September 22nd, 2004

A 3D ANALOGUE OF PYTHAGORAS' THEOREM
PRESENTED BY
PETER WAKEFIELD SAULT

Copyright © Peter Wakefield Sault 1976-2004

INTRODUCTION

The reader is surely already familiar with Pythagoras' 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.

Figure 1.

The application of Pythagoras' 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.

Figure 2.

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' 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.

A 3D ANALOGUE OF PYTHAGORAS' THEOREM

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.

Figure 3.

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

Figure 4.

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) - 4y⁴
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.

Figure 5.

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.

A 3D ANALOGUE of PYTHAGORAS' THEOREM