## Wednesday, March 29, 2006

### Equiangular Triangles

In today's blog, I am reviewing the background to the concepts of sin and cosin. The major assumption behind sin and cosin is that the ratios of the sides of right triangles can be calculated based solely on the measurement of an angle. In other words, the ratio between sides ( opposite side over hypotenuse for sin and adjacent side over hypotenuse for cosin) is constant for all similar right triangles.

Two right triangles are similar if they share the same angles but not may not share the same sides. The important idea that is presented in Euclid's Elements is that the ratio of two sides is equal to the ratio of two corresponding sides for any equiangular triangle. This property of similar triangles is enough to show that sin and cosin depend solely on the measurement of the angle. I will talk more about this in a future blog.

I will only go over enough which are necessary to establish the corresponding sides property of similar triangles.

Lemma 1: If two triangles have the same height, then the ratio of their areas is equal to a ratio of their bases.

Proof:

(1) Let a1 be the area of triangle ABC with height h and base b1

(2) Let a2 be the area of triangle DEF with height h and base b2

(3) Now a1/a2 = [(1/2)b1h]/[(1/2)b2h] = b1/b2 [See Lemma 2, here]

QED

Lemma 2: If a parallel line cuts through a triangle, it divides the sides of the triangle proportionally.

Proof:

(1) Let DE be a parallel line that cuts through the triangle ABC

(2) We can see that the areas of triangle DEB and triangle DEC are equal since:

(a) They share the same base DE

(b) They have the same height [Based on Lemma 2, here]

(3) Since triangle DEB and triangle DEC have the same area, we know that:

(area DEB)/(area ADE) = (area DEC)/(area ADE)

(4) Now triangle DEB and triangle ADE have the same height.

(5) So, we also know that from Lemma 1 above:
(area DEB)/(area ADE) = DB/AD

(6) Likewise, triangle AED and triangle DEC have the same height so Lemma 1 gives us:
(area DEC)/(area ADE) = EC/AE

(7) Putting this all together (steps #3, #5, and #6) gives us:
DB/AD = EC/AE

QED

Postulate 1: Parallel Postulate

If two lines intersect the same line and the sum of their intersectings angles is less than 180 degrees, these lines will eventually intersect.

For more details on the Parallel Postulate, see here. This is the most famous postulate of all Euclid's Elements.

Lemma 3: If two triangles are equiangular, then the sides about equal angles are proportional where the corresponding sides are opposite the equal angles.

Proof:

(1) Let ABC and DCE be equiangular triangles.

(2) Let us assume that BC and CE are colinear.

(3) Since ∠ ABC ≅ ∠ DCE, we know that FB is parallel to DC [See here for definition of parallel lines]

(4) Since ∠ BCA ≅ ∠ CED, we know that AC is parallel to FE [See here for definition of parallel lines]

(5) By the Parallel Postulate above, we know that if we extend line AB and line DE, they will intersect at a point F.

(6) Now, from (3) and (4), ACDF is a parallelogram [See here for definition of a parallelogram]

(7) Therefore FA ≅ DC and AC ≅ FD [See Lemma 1 here]

(8) Since AC is a parallel line that cuts through triangle FBE, Lemma 2 above gives us:
BA/AF = BC/CE

which means that

BA * CE = BC * AF

and further that:

BA/BC = AF/CE

(9) Since AF ≅ DC [from #7], we have
BA/BC = DC/CE

(10) Since DC is a parallel line that also cuts through triangle FBE, we get:
BC/CE = FD/DE

(11) Now combining (#7) and (#9), we get:
BC/CE = FD/DE = AC/DE

which means that:

BC * DE = AC * CE

and further that:

BC/AC = CE/DE

(12) Combining (#11) and (#9) gives us:

BA/BC = DC/CE [from #9]
BC/AC = CE/DE [from #11]

So that we have:

BA*CE = BC*DC
BC*DE = AC*CE

And we have:

BA*CE*BC*DE = BC*DC*AC*CE

So that we can divide out CE and BC to get:

BA*DE = AC*DC

And finally that:

BA/AC = DC/DE

(13) So we are done since we have:
BA/BC = DC/CE [Step #9]
BC/AC = CE/DE [Step #11]
BA/AC = DC/DE [Step #12]

QED

References