triangle – 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 44

To a given straight line to apply, in a given rectilineal angle, a parallelogram equal to a given triangle.
[Desmos graphs can be found here]

Let AB be the given straight line, C be the given triangle, and D the given angle.

We will apply a parallelogram to AB in an angle equal to D that is equal to the triangle C.
Use Proposition 42 (covers how to construct a parallelogram, in a given angle, that is equal to a given triangle) to construct the parallelogram BEFG that is equal to the triangle C in the angle EBG that is equal to D. Place the parallelogram so that BE is in a straight line with AB.

Let FG be drawn through to a point H such that AH is parallel to BG and EF. (Use Proposition 31: through a given point to draw a straight line that is parallel to a given line.)

Join HB.

Since the straight line HF falls on the parallel lines AH and EF, Proposition 29 (a straight line falling on two parallel lines creates two interior angles on the same side that are equal to two right angles) tells us that the angles AHF and HFE together are equal to two right angles.
Therefore the two angles BHG and GFE are less than two right angles.
(BHG is contained in and therefore less than AHF, and angle GFE is equal to HFE.)
Straight lines produced indefinitely from angles less than two right angles must meet, so the line through HB must meet the line through FE.
Let HB and FE be produced and meet at a point K.
Through K use Proposition 31 to draw a line KL that is parallel to FH and EA.
Produce HA to the point L and produce GB to a point M that lies on LK.

HLKF is a parallelogram, and HK is its diameter.
ABGH and MKEB are parallelograms contained in HLKF.
LMBA and BEFG are the complements about HK, and by Proposition 43 (in any parallelogram the complements of the parallelograms about the diameter are equal to one another) the two parallelograms LMBA and BEFG are equal.

Parallelogram BEFG was constructed to be equal to triangle C, so parallelogram LMBA is also equal to the triangle C.

Angle ABM is equal to angle GBE by Proposition 15 (if two straight lines intersect each other, they make the vertical angles equal to one another), and since angle GBE was constructed equal to angle D then the angle ABM is equal to angle of D.

Therefore parallelogram LMBA equal to the given triangle C has been applied to the given straight line AB, in the angle ABM equal to the given angle D.

The next proposition involves constructing a parallelogram in a given angle that is equal to a given rectilineal figure (instead of a given triangle) – George

Book 1 – Proposition 42

To construct, in a given rectilineal angle, a parallelogram equal to a given triangle.
[Desmos graphs can be found here]

Start with a given triangle ABC and a given rectilineal angle D.

The goal is to construct a parallelogram that is equal to the triangle ABC in an angle equal to D.

Let E bisect BC. Join AE. Use Proposition 23 (Construction of an angle on a straight line that is equal to a given angle) to construct an angle CEF that is equal to angle D.

Draw AG through A that is parallel to EC. Draw CG through C that is parallel to EF. These lines can be drawn using Proposition 31 (Construction of a straight line through a given point that is parallel to a given line).

Since FG is parallel to EC and EF is parallel to CG, FECG is a parallelogram.

Now we look at triangles ABE and AEC.
Since BE is equal to EC (as E bisects BC), the two triangles have equal bases.
The two triangles are in the same parallels (BC and AG).
By Proposition 38 (two triangles on equal bases and in the same parallels are equal to one another) triangle ABE is equal to triangle AEC.

Since the triangle ABC is composed of the two equal triangles ABE and AEC, the triangle ABC is double the triangle AEC.

But, by Proposition 41 (if a parallelogram has the same base as a triangle and they are in the same parallels then the parallelogram is double the triangle) the parallelogram FECG is double the triangle AEC since they share the same base (EC) and they are in the same parallels.

Since both the parallelogram FECG and the triangle ABC are double the triangle AEC, the parallelogram FECG is equal to the triangle ABC.
The angle FEC is equal to the angle D.
We have constructed the parallelogram that meets the necessary conditions.

Book 1 – Proposition 41

If a parallelogram has the same base as a triangle and is in the same parallels, then the parallelogram is double the triangle.
[Desmos graphs can be found here]

We begin with parallelogram ABCD and triangle EBC sharing the same base BC and be in the same parallels BC and AE.

We will prove that the parallelogram ABCD is double the triangle EBC.
Start by joining AC.

Compare triangles ABC and EBC. They have the same base BC, and they are in the same parallels BC and AE, so the two triangles are equal by Proposition 37. (Triangles with the same base that are in the same parallels are equal to one another.)

Since AC is the diameter of the parallelogram ABCD, the parallelogram is double the triangle ABC by Proposition 34. (The diameter of a parallelogram bisects it.)

Since the parallelogram ABCD is double the triangle ABC, and the triangle ABC is equal to the triangle EBC, therefore the parallelogram ABCD is double the triangle EBC.

Q.E.D.

In the next proposition we return to a construction of a parallelogram that is equal to a given triangle – George

Book 1 – Proposition 40

Equal triangles which are on equal bases and on the same side are also in the same parallels.
[Desmos graphs can be found here]

Let ABC and CDE be equal triangles with equal bases BC and CE.

We will prove that AD is parallel to BE to show that the two triangles are in the same parallels.
Begin by joining AD.

If AD is not parallel to BE, then draw AF through point A so that AF is parallel to BE. (Proposition 31 covers drawing a line through a given point that is parallel to a given line.) Join FE as well.

Triangles ABC and FCE have equal bases (BC and CE) and in the same parallels (BE and AF), so by Proposition 38 (triangles on equal bases and in the same parallels are equal to one another) triangles ABC and FCE are equal to one another.
Since triangle ABC is also equal to triangle CDE, triangle FCE must also be equal to triangle CDE which is impossible. (One of those triangles must be greater than the other.)

Therefore AF is not parallel to BE.

Therefore AD is parallel to BE, and the two triangles are in the same parallels.

Q.E.D.

The next proposition will compare triangles and parallelograms with equal bases in the same parallels – George

Book 1 – Proposition 39

Equal triangles which are on the same base and on the same side are also in the same parallels.
[Desmos graphs can be found here]

Let triangles ABC and DBC have the same base BC and both be on the same side of it.

Join AD. We will prove that AD is parallel to BC.

If AD is not parallel to BC, let A be joined to a point E on BD such that AE is parallel to BC. (Proposition 31 covers drawing a line through a given point that is parallel to a given line.) Join EC.

By Proposition 37 (triangles on the same base and in the same parallels are equal) triangle ABC is equal to triangle EBC as they share the same base (BC) and are in the same parallels.

Since triangle ABC is equal to triangle DBC, then triangle EBC is also equal to triangle DBC which is impossible.
Therefore AE cannot be parallel to BC.
The same is true for any other straight line except AD.
Therefore AD is parallel to BC.

Q.E.D.

In the next proposition we will extend “the same base” to “equal bases” – George

Book 1 – Proposition 38

Triangles which are on equal bases and in the same parallels are equal to one another.
[Desmos graphs can be found here]

Let triangles ABC and DEF be on equal bases BC and EF, and in the same parallels BF and AD.

We will prove that triangle ABC is equal to triangle DEF.
Let AD be produced in both direction to points G and H such that BG is parallel to CA and FH is parallel to ED. (Proposition 31 covers drawing a line through a given point that is parallel to a given line.)

The parallelogram GBCA is equal to the parallelogram DEFH by Proposition 36 as they are on equal bases BC and EF and are in the same parallels BF and GH.
(Proposition 36 states that two parallelograms on equal bases that are in the same parallels are equal to each other.)

The diameter AB bisects the parallelogram GBCA, so the triangle ABC is half of the parallelogram GBCA by Proposition 34 (the diameter of a parallelogram bisects the areas).
By similar reasoning, Proposition 34 tells us that triangle DEF is half of the parallelogram DEFH since DF is a diameter of the parallelogram.

Triangles ABC and DEF are each half of two equal parallelograms, and therefore are equal to each other.

Q.E.D.

Book 1 – Proposition 37

Triangles which are on the same base and in the same parallels are equal to one another.
[Desmos graphs can be found here]

Let ABC and DBC be triangles on the same base BC and in the same parallels AD and BC.

We will prove that triangles ABC and DBC are equal.

Project the line AD in both directions to points E and F such that BE is parallel to CA and CF is parallel to BA.
Proposition 31 covered how to construct a line through a given point that is parallel to a given line.
(BE passes through B and is parallel to CA. CF passes through C and is parallel to BA.)

EBCA and DBCF are parallelograms. The two parallelograms share the same base (BC) and are in the same parallels (BC and EF), so by Proposition 35 they are equal.

By Proposition 34 (the diameter of a parallelogram bisects its area) triangle ABC is half of the parallelogram EBCA because its diameter AB bisects the parallelogram.

Again by Proposition 34, since DC is the diameter of the parallelogram DBCF, triangle DBC is half of the parallelogram DBCF.

Since the triangles ABC and DBC are each half of equal parallelograms EBCA and DBCF, the two triangles ABC and DBC are therefore equal to one another.

Q.E.D.

The next proposition extends Proposition 37 from “the same base” to “equal bases” – George

Book 1 – Proposition 32

In any triangle, if one of the sides be produced, the exterior angle is equal to the two interior and opposite angles, and the three interior angles of the triangle are equal to two right angles.
[Desmos graphs can be found here]

Start with triangle ABC, and let side BC be produced to D.

We will prove that (i) the exterior angle ACD is equal to the two interior and opposite angles ABC and CAB, and (ii) the three interior angles ABC, BCA, and CAB are equal to two right angles.

Using Proposition 31 (which covered how to construct a line through a given point that is parallel to a given line), let CE be drawn through point C that is parallel to AB.

Line AC falls upon the two parallel lines AB and CE, so by Proposition 29 (a straight line falling on two parallel lines makes the alternate angles equal to one another) the alternate angles BAC and ACE are equal to one another.

Line BD also falls upon the two parallel lines AB and CE, so Proposition 29 (exterior angle is equal to interior and opposite angle) tells us that the exterior angle ECD is equal to the opposite and interior angle ABC.

Angle ACD is made up of the angles ACE (which is equal to angle BAC) and ECD (which is equal to angle ABC). So, the exterior angle ACD is equal to the two interior and opposite angles ABC and BAC.

Moving on to the proof of (ii), add the angle ACB to both angle ACD as well as to the angles ABC and BAC.
Angles ACD and ACB are equal to the angles ABC, BAC, and ACB.
By Proposition 13 (a straight line set up on a straight line makes two angles that are equal to two right angles) angles ACD and BAC are equal to two right angles.

Therefore the three interior angles ABC, BAC, and ACB are equal to two right angles.

Q.E.D.

That’s two-thirds of the way through Book 1. Comments or questions? I’d love to hear them – George