帕普斯对毕达哥拉斯定理的推广Pappus's extension of the Pythagorean Theorem

89° Pappus's extension of the Pythagorean Theorem. [The following is adapted, with permission, from the article, by Howard Eves, of the same title that appeared in the Historically Speaking section of The Mathematics Teacher, November, 1958, pp. 544–546.]

Every student of high school geometry sooner or later becomes familiar with the famous Pythagorean Theorem, which states that in a right triangle the area of the square described on the hypotenuse is equal to the sum of the areas of the squares described on the two legs. This theorem appears as Proposition 47 in Book I of Euclid's Elements, written about 300 B.C.

Even in Euclid's time, certain generalizations of the Pythagorean Theorem were known. For example, Proposition 31 of Book VI of the Elements states: In a right triangle the area of a figure described on the hypotenuse is equal to the sum of the areas of similar figures similarly described on the two legs. This generalization merely replaced the three squares on the three sides of the right triangle by any three similar and similarly described figures. A more worthy generalization stems from Propositions 12 and 13 of Book II. A combined and somewhat modernized statement of these two propositions is: In a triangle, the square of the side opposite an obtuse (acute) angle is equal to the sum of the squares on the other two sides increased (decreased) by twice the product of one of these sides and the projection of the other side on it. That is, in the notation of Figure 13, $(AB)^2 = (BC)^2 + (CA)^2 \pm 2(BC)(DC)$, the plus or minus sign being taken according as angle $C$ of triangle $ABC$ is obtuse or acute.

(Figure 13: two triangles, illustrating the obtuse case (left) and the acute case (right), with $AD \perp BC$ in each.)

If we employ directed line segments we may combine Propositions 12 and 13 of Book II and Proposition 47 of Book I into the single statement: If in triangle $ABC$, $D$ is the foot of the altitude on side $BC$, then $(AB)^2 = (BC)^2 + (CA)^2 - 2(BC)(DC)$. Since $DC = CA \cos BCA$, we recognize this last statement as essentially the so-called law of cosines, and the law of cosines is indeed a fine generalization of the Pythagorean Theorem.

But perhaps the most remarkable extension of the Pythagorean Theorem that dates back to the days of Greek antiquity is that given by Pappus of Alexandria at the start of Book IV of his Mathematical Collection. The Pappus extension of the Pythagorean Theorem is as follows (see Figure 14): Let $ABC$ be any triangle and $CADE$, $CBFG$ any parallelograms described externally on sides $CA$ and $CB$. Let $DE$ and $FG$ meet in $H$ and draw $AL$ and $BM$ equal and parallel to $HC$. Then the area of parallelogram $ABML$ is equal to the sum of the areas of parallelograms $CADE$ and $CBFG$. The proof is easy, for we have $CADE = CAUH = SLAR$ and $CBFG = CBVH = SMBR$. Hence $CADE + CBFG = SLAR + SMBR = ABML$. It is to be noted that the Pythagorean Theorem has been generalized in two directions, for the right triangle in the Pythagorean Theorem has been replaced by any triangle, and the squares on the legs of the right triangle have been replaced by any parallelograms.

(Figure 14: parallelograms $CADE$ and $CBFG$ constructed externally on sides $CA$ and $CB$ of $\triangle ABC$, with $DE$ and $FG$ meeting at $H$, and parallelogram $ABML$ drawn on $AB$ with $AL \parallel BM \parallel HC$.)

The student of high school geometry can hardly fail to be interested in the Pappus extension of the Pythagorean Theorem, and the proof of the extension can serve as a nice exercise for the student. Perhaps the more gifted student of geometry might like to try his hand at establishing the further extension (to three-space) of the Pappus extension: Let $ABCD$ (see Figure 15) be any tetrahedron and let $ABD\text{-}EFG$, $BCD\text{-}HIJ$, $CAD\text{-}KLM$ be any three triangular prisms described externally on the faces $ABD$, $BCD$, $CAD$ of $ABCD$. Let $Q$ be the point of intersection of the planes $EFG$, $HIJ$, $KLM$, and let $ABC\text{-}NOP$ be the triangular prism whose edges $AN$, $BO$, $CP$ are translates of the vector $QD$. Then the volume of $ABC\text{-}NOP$ is equal to the sum of the volumes of $ABD\text{-}EFG$, $BCD\text{-}HIJ$, $CAD\text{-}KLM$. A proof analogous to the one given above for the Pappus extension can be supplied.

(Figure 15: tetrahedron $ABCD$ with three external triangular prisms $ABD\text{-}EFG$, $BCD\text{-}HIJ$, $CAD\text{-}KLM$, and the resulting prism $ABC\text{-}NOP$.)