Proposition 5
Theorem
If a line AB is divided into two equal parts at point C, and also into two unequal parts at point D, then ￜADￜ⋅ￜDBￜ + ￜCDￜ2 = ￜCBￜ2 .
Under Development
Last updated on: 2022-09-19.
Commentary
1. Let AB be the given line segment. Let C be the midpoint of AB , and let D be any other point on AB .
2. Construct a square◇CBEF on BC , and find a point G on EF such that DG is parallel to BE and intersects the diagonal BF at a point H.
3. Through H, draw a line parallel toAB and intersecting CF and BE at L and M, respectively.
4. Construct a lineAK that is perpendicular to AB , with K being the intersection with the line LM that is parallel to AB .
5. The rectangle◇ACLK has been constructed with the area equal to that of rectangle ◇DBEG .
6. The sum of the areas of rectangle◇ADHK (ￜADￜ⋅ￜDBￜ ) and square ◇LHGF (ￜCDￜ2 ) is equal to the area of square ◇CBEF (ￜCBￜ2 ).
7. This geometric relationship is similar to the next proposition, Book 2 Proposition 6, and can be expressed algebraically as(x + y) (x - y) + y2 = x2 (where ￜACￜ = x and ￜCDￜ = y ). This algebraic relationship is one of the most used polynomial identities.
2. Construct a square
3. Through H, draw a line parallel to
4. Construct a line
5. The rectangle
6. The sum of the areas of rectangle
7. This geometric relationship is similar to the next proposition, Book 2 Proposition 6, and can be expressed algebraically as
Original statement
ἐὰν ϵὐθϵῖα γραμμὴ τμηθῇ ϵἰς ἴσα καὶ ἄνισα, τὸ ὑπὸ τῶν ἀνίσων τῆς ὅλης τμημάτων πϵριϵχόμϵνον ὀρθογώνιον μϵτὰ τοῦ ἀπὸ τῆς μϵταξὺ τῶν τομῶν τϵτραγώνου ἴσον ἐστὶ τῷ ἀπὸ τῆς ἡμισϵίας τϵτραγώνῳ.
English translation
If a straight line is cut into equal and unequal segments, the rectangle contained by the unequal segments of the whole together with the square on the straight line between the points of section is equal to the square on the half.
