P l a n e G e o m e t r y
An Adventure in Language and Logic
|1.||From a given point to draw a straight line equal to a given straight line.|
|2.||Let A be the given point, and BC the given straight line;|
|3.||we are required to draw from the point A a straight line equal|
|(The construction begins by drawing the equilateral triangle ABD; then extending DA, DB, to E and F; drawing a circle with radius BC,
thus making BG equal to BC; then making DG, DL radii of a circle, so that on subtracting DB, DA respectively, AL will equal BG and therefore BC.)
|5.||From the point A to the point B, draw the straight line AB;|
|6.||and on it draw the equilateral triangle ABD;||(I. 1)|
|7.||and extend the straight lines DA, DB to E, F.||(Postulate 2)|
|8.||With B as center and BC as radius, draw the circle CHG,|
|9.||meeting DF at G.||(Postulate 3)|
|10.||With D as center and DG as radius, draw the circle GKL,|
|11.||meeting DE at L.||(Postulate 3)|
|12.||Then AL will equal BC.|
|13.||For, since B is the center of the circle CHG,|
|14.||BC is equal to BG.||(Definition 16)|
|15.||And since D is the center of circle GKL,|
|16.||DL is equal to DG.|
|17.||But in those lines, DA is equal to DB;||(Definition 9)|
|18.||therefore the remainder AL is equal to the remainder BG.|
|19.||And we have shown that BC is equal to BG;|
|20.||therefore each of the straight lines AL, BC is equal to BG.|
|21.||And things equal to the same thing are equal to one another.|
|22.||Therefore AL is equal to BC.|
|23.||Therefore from the given point A we have drawn a straight line|
|24.||AL equal to the straight line BC.|
|25.||Which is what we wanted to do.|
This proposition leads directly to the next one, where we will be required to cut off from the longer of two straight lines a length equal to the shorter line. The solution is obvious -- but notice how we must rely on Proposition 2; line 6 below. In fact, the next proposition is the only one that requires Proposition 2. The "given" point is the endpoint of a line.
The student should now begin to see how each proposition depends on previous propositions. That is the nature of any logical theory. That is the axiomatic method.
|1.||Given two unequal straight lines, to cut off from the longer line|
|2.||a straight line equal to the shorter line.|
|3.||Let AB and C be the two given straight lines, and let AB be|
|5.||we are required to cut off from AB a straight line equal to C.|
|6.||From the point A draw AD equal to C;||(Proposition 2)|
|7.||and with A as center and radius AD, draw the circle DEF.|
|8.||Then, since the point A is the center of circle DEF,|
|9.||AE is equal to AD.||(Definition 16)|
|10.||But C is also equal to AD.||(Construction)|
|11.||Therefore each of the lines AE, C is equal to AD.|
|12.||Therefore AE is equal to C.||(Axiom 1)|
|13.||Therefore from AB the longer of two straight lines|
|14.||we have cut off a straight line AE equal to C, the shorter line.|
|15.||Which is what we wanted to do.|
Please "turn" the page and do some Problems.
Continue on to the next proposition.
Please make a donation to keep TheMathPage online.
Even $1 will help.
Copyright © 2015 Lawrence Spector
Questions or comments?