## Friday, April 21, 2006

### More on Equiangular Triangles

In today's blog, I review more proofs on equiangular triangles that are taken straight from Euclid's Elements. I use these properties in showing that the ratio of circumference to diameter for any circle is always the same (which is, of course, the definition of pi).

Lemma 1: If two triangles have one angle equal to one angle and the sides around the equal angle proportional, then the two triangles are equiangular.

Proof:

(1) Assume that ∠ BAC ≅ ∠ EDF and BA/AC = ED/DF

(2) There exists a point G such that ∠ FDG ≅ ∠ BAC and ∠ DFG ≅ ∠ ACB [See here for details on the construction.]

(3) ∠ B ≅ ∠ G [since the angles of a triangle add up to 180 degrees, see Lemma 4 here for details]

(4) triangle ABC is equiangular with triangle DGF [from step #2 and step #3]

(5) From the properties of equiangular triangles (see here), we know that:

BA/AC = GD/DF

(6) From our assumption in Step #1, we can conclude that:

GD/DF = ED/DF

(7) And from step #6, we can conclude that:

ED ≅ GD

(8) From Step #1 and Step #2, we can conclude that:

∠ EDF ≅ FDG

(9) We can now conclude that triangle DEF ≅ triangle DGF from Side-Angle-Side (see Postulate 1 here) since:

(a) DF ≅ DF

(b) ∠ EDF ≅ ∠ FDG [Step #8]

(c) ED ≅ GD [Step #7]

(10) ∠ ACB ≅ ∠ DFE since:

(a) ∠ ACB ≅ ∠ DFG [Step #2]

(b) ∠ DFG ≅ ∠ DFE [From Step #9, see here for properties of congruent triangles if needed]

(11) Finally, we can conclude that triangle ABC is equiangular with triangle DEF since:

(a) ∠ BAC ≅ ∠ EDF [By assumption in Step #1]

(b) ∠ ACB ≅ ∠ DFE [By Step #10]

(c) ∠ B ≅ ∠ E [Since angles of triangles add up to 180 degrees]

QED

Lemma 2: Those triangles which have one angle equal to one angle and which the sides about the equal angles are reciprocally proportional, are equal.

Proof:

(1) Let the sides of triangle ABC and triangle ADE be reciprocally proportional so that:

(2) Since triangle ABC and triangle ABD share the same height, we can conclude that (see here):

AC/AD = area triangle ABC/area triangle ABD

(3) Likewise, since triangle ADE and triangle ABD share the same height, we can conclude that:

AE/AB = area triangle ADE/triangle ABD

(4) Therefore, triangle ABC/triangle ABD = triangle ADE/triangle ABD.

(5) But then we have:

area triangle ABC * area triangle ABD = area triangle ADE * area triangle ABD

(6) And if we divide both sides by the area of triangle ABD, we get:

area triangle ABC = area triangle ADE.

QED

Lemma 3: Similar triangles are one to another in the square ratio of corresponding sides.

Proof:

(1) Let triangle ABC and triangle DEF be equiangular triangles such that:

∠ A ≅ ∠ D
∠ B ≅ ∠ E
∠ C ≅ ∠ F

(2) There exists a point G such that BG = [(EF)*(EF)]/BC

(3) From (#2) we have:

1/BG = BC/[(EF)*(EF)]

which implies that:

EF/BG = BC/EF

(4) From a Property of Equiangular Triangles (see here if needed), we know that:

AB/BC = DE/EF

so that:

AB/DE = BC/EF

(5) We can also conclude that triangle ABG has equal area to triangle DEF since:

(a) B ≅ ∠ E (step #1)

(b) AB/DE = EF/BG (from combining step #3 with step #4)

(c) Lemma 2 above

(6) We can also conclude that BC/BG = (CB)2/(EF)2 since:

BC*BG = EF*EF (step #2)

And if we multiply BC to both sides, we get:

(BC)2*BG = (EF)2*BC

If we divide BG from both sides and (EF)2 from both sides, we get:

(BC)2/(EF)2 = BC/BG

(7) Since triangle ABC and triangle ABG have the same height, we can conclude (see here):

CB/BG = area triangle ABC/area triangle ABG

(8) Applying setp #6, gives us:

area triangle ABC/area triangle ABG = (BC)2/(EF)2

(9) Finally, from step #5, we have:

area triangle ABC/area triangle DEF = (CB)2/(EF)2

QED

References