right angles – uCLID

Book 1 – Proposition 48

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

Book 1 – 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.
[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

Book 1 – Proposition 11

Proposition 11

To draw a straight line at right angles to a given straight line from a given point on it.

[Desmos graphs are located here]

We will start with a line segment AB, and let C be the given point on it.

We will draw a line from point C at right angles from the segment AB.
We begin by plotting a point D on AC.
Let E be the point on CB such that CE is equal to CD. (Proposition 3 states that we can cut off a line segment that is equal to CD.)

On DE we can construct the equilateral triangle FDE (Proposition 1 shows the construction that shows an equilateral triangle can be built on any line segment), and join points F and C as shown.

We will now show that FC forms right angles with AB at point C by looking at triangles DCF and ECF.

DC is equal to EC by the way we chose point E.
CF is common to both triangles.
DF is equal to FE because they are both sides of the equilateral triangle DFE.

Therefore by Proposition 8 (if two triangles have three equal sides then the angles between two equal sides are equal), the angles DCF and ECF are equal to each other.

One of Euclid’s definitions will help us here: When a straight line set up on a straight line makes the adjacent angles equal to one another, each of the angles is right.

The two adjacent angles DCF and ECF are formed when a straight line (CF) is set up on a straight line (AB) and are equal to each other, so each of the angles is right.

This proposition will be used in the next proposition, a construction of drawing a line through a point that is perpendicular to a given straight line.

Questions, comments, or feedback? Please leave a comment below – George