Proposition 1
Theorem
If AB and CD are line segments and E is a point on CD , then ￜAB ￜ⋅ￜCD ￜ = ￜAB ￜ⋅ￜCE ￜ + ￜAB ￜ⋅ￜED ￜ .
Commentary
1. Let AB and CD be the given line segments, and let E be any point on CD .
2. Construct the rectangle◇CDPQ on CD , with ￜCQ ￜ = ￜAC ￜ .
3. Pick a point H onPQ and draw EH perpendicular to PQ so that EH has the same length as AB and EH is a shared side of ◇CEHQ and ◇EDPH .
4. The area of the rectangle◇CDPQ (ￜAB ￜ⋅ￜCD ￜ ) is the sum of the areas of the two rectangles ◇CEHQ (ￜAB ￜ⋅ￜCE ￜ ) and ◇EDPH (ￜAB ￜ⋅ￜED ￜ ).
5. This geometric relationship can be expressed algebraically as follows: ify = a + b , then x y = x a + x b (where ￜAB ￜ = x , ￜCD ￜ = y , ￜCE ￜ = a and ￜED ￜ = b ). This algebraic relationship is known as the distributive law.
6. Euclid stated this proposition to be true for any number of pieces of one of the given line segments, while for demonstration purposes, we have chosen the case for two pieces. The general algebraic relationship can be represented as: ify = a1 + a2 + … + an then x y = x a1 + x a2 + … + x an .
2. Construct the rectangle
3. Pick a point H on
4. The area of the rectangle
5. This geometric relationship can be expressed algebraically as follows: if
6. Euclid stated this proposition to be true for any number of pieces of one of the given line segments, while for demonstration purposes, we have chosen the case for two pieces. The general algebraic relationship can be represented as: if
Original statement
ἐὰν ὦσι δύο ϵὐθϵῖαι, τμηθῇ δὲ ἡ ἑτέρα αὐτῶν ϵἰς ὁσαδηποτοῦν τμήματα, τὸ πϵριϵχόμϵνον ὀρθογώνιον ὑπὸ τῶν δύο ϵὐθϵιῶν ἴσον ἐστὶ τοῖς ὑπό τϵ τῆς ἀτμήτου καὶ ἑκάστου τῶν τμημάτων πϵριϵχομένοις ὀρθογωνίοις.
English translation
If there are two straight lines, and one of them is cut into any number of segments whatever, the rectangle contained by the two straight lines is equal to the rectangles contained by the uncut straight line and each of the segments.
