Now that I have made it to the end of Book 1, I am going to take a short break before moving on to Book 2. I will use that time to add comments and insights to the 48 Propositions in Book 1.
If you have any comments on things that could be added to those first 48 please leave me a comment.
You can follow me @uCLID_blog on Twitter to keep up with my progress and be notified when I move on to Book 2.
Thanks – George
If in a triangle the square one one of the sides is equal to the squares on the remaining two sides of the triangle, the angle contained by the remaining two sides of the triangle is right.
[Desmos graphs can be found here]
For the triangle ABC let the square on side BC be equal to the squares on side AB and BC.
We will prove that angle BAC is a right angle.
Let AD be drawn from point A at a right angle to AC.
Let AD be made equal to AB.
Join DC.
Since DA is equal to AB, the square on DA is equal to the square on AB.
Add the square on AC to each.
The sum of the squares on DA and AC is equal to the sum of the squares on AB and AC.
The square on DC is also equal to the sum of the squares on DA and AC because the angle DAC is a right angle. (Proposition 47: In right-angled triangles the square on the side subtending the right angle is equal to the squares on the sides containing the right angle.)
Since the square on BC is equal to the squares on AB and AC, the square on BC is also equal to the square on DC. Therefore the side DC is equal to the side BC.
Now compare triangle DAC to triangle BAC.
Side DA was constructed to be equal to side AB, side AC is common, and the two bases DC and BC were shown to be equal.
Therefore by Proposition 8 (side-side-side) triangle DAC is equal to triangle BAC and angle DAC is equal to angle BAC.
Since angle DAC is a right angle, therefore angle BAC is a right angle.
Q.E.D.
Proposition 47 showed that for a right triangle that the square on the side subtending the right angle is equal to the squares on the sides containing that right angle. Proposition 48 takes us in a different direction. If the square on one side is equal to the squares of the other two sides, then the angle opposite that one side must be a right angle. Notice how the two propositions work in the opposite direction of each other.
We have reached the end of Book 1 of Euclid’s Elements. After a short break I will continue on with Book 2. Thanks for reading – George
In right-angled triangles the square on the side subtending the right angle is equal to the squares on the sides containing the right angle.
[Desmos graphs can be found here]
Let ABC be a right-angled triangle with angle BAC being a right angle.
We will prove that the square on BC is equal to the squares on AB and AC.
Use Proposition 46 (the construction of a square on a straight line) to construct the square BDEC on the line BC, the square AGFB on the line AB, and the square AHKC on the line AC.
Through the point A draw the line AL that is parallel to BD and CE.
Join AD and CF.
Angles BAC and BAG are each right angles. The straight line BA falls on the two straight lines AC and AG on different sides of it. The adjacent angles are equal to two right angles, so GA is in a straight line with AC by Proposition 14 (if any line and any two lines not on the same side of it form two adjacent angles that are equal to two right angles, then the two lines will be in a straight line with one another).
Also. by the same reasoning, BA is in a straight line with AH.
The right angle DBC is equal to the right angle FBA.
Add angle ABC to each. Therefore angle DBA is equal to angle FBC.
SIde DB is equal to side BC (on the same square), and side AB is equal to side FB.
Comparing triangles ABD and FBC: Sides AB and BD are equal to sides FB and BC and the angles between the two sides are also equal. So, by Proposition 4 (side-angle-side) the base AD is equal to the base FC and the triangle ABD is equal to the triangle FBC.
By Proposition 41 (if a parallelogram and a triangle share the same base and are in the same parallels, then the parallelogram is double the triangle) the parallelogram BL is double the triangle ABD.
Also by Proposition 41, the square AGFB is double the triangle FBC as they share the same base (FB) and are in the same parallels (FB and GC).
Since the parallelogram BL is double the triangle ABD and the square AGFB is double the triangle FBC, and those two triangles are equal to each other, the parallelogram BL is equal to the square AGFB.
Join AE and BK.
In a similar fashion we can prove that the parallelogram CL is equal to the square AHKC.
The whole square BDEC is equal to the parallelograms BL and CL, which are equal to the two squares AGFB and AHKC.
The square BDEC is described on BC, the square AGFB is described on AB, and the square AHKC is described on AC.
Therefore the square on BC (subtending the right angle) is equal to the squares on AB and AC (the sides containing the right angle).
Q.E.D.
This Proposition is equivalent to the Pythagorean theorem. If the lengths of the two legs of a triangle are a and b and the hypotenuse is c, then a^2 + b^2 = c^2. In this Proposition and proof you can actually see the squares instead of working with an abstract formula. There are many proofs of the Pythagorean theorem – I encourage you to look up some of the others.
George
On a given straight line to describe a square.
[Desmos graphs can be found here]
Let AB be the given straight line. We will construct a square on AB.
Use Proposition 11 (constructing a straight line at a right angle to a given line at a point) to draw AC at a right angle to AB at the point A.
Select point D on AC such that AD is equal to AB.
Using Proposition 31 (covers drawing a straight line through a point that is parallel to a given line), through point D draw DE that is parallel to AB.
Also, through point B draw BE that is parallel to AD.
ADEB is a parallelogram, and by Proposition 34 (in a parallelogram opposite sides are equal and opposite angles are equal) AB is equal to DE and AD is equal to BE.
Since AB is equal to AD, all four sides of the parallelogram are equal to each other. Therefore the parallelogram ADEB is equilateral.
Now we will show that the parallelogram is also right-angled, making it a square.
Since AD falls upon the parallel lines AB and DE, Proposition 29 (if a line falls on two parallel lines, then the interior angles on the same side are equal to two right angles) tells us that the two angles BAD and ADE are together equal to two right angles.
But angle BAD is a right angle, so angle ADE is also a right angle.
Using Proposition 34 again, the two angles opposite BAD and ADE are equal to those two angles. So, angles ABE and BED are also each right angles.
Since ADEB is right-angles and equilateral, it is a square described on the straight line AB.
Next proposition … the Pythagorean Theorem!