Thursday, March 23, 2006


In today's blog, I review some basic proofs from Euclid's Elements relating to Parallelograms. These extend the results on parallel lines and are needed for the proofs on similar triangles which I use in my discussion about sin and cosin.

The diagrams are taken from David Joyce's web site on Euclid's Elements which I highly recommend.

Definition: Parallelogram

A parallelogram is any four-sided shape where opposite sides are parallel to each other.

Lemma 1: In parallelograms, opposite sides and opposite angles are congruent.


(1) Let ABCD be a parallelogram.

(2) AD is parallel to BC [Definition of Parallelogram]

(3) Alternate angles are congruent [see Lemma 2 here] gives us:
∠ DAC ≅ ∠ BCA
∠ DCA ≅ ∠ BAC

(4) Since line AC is congruent to itself, we can use the ASA lemma (see here) to conclude that triangle DAC ≅ triangle BCA

(5) But then corresponding sides are congruent which gives us (see here for definition of Congruent Triangles):

(6) And opposite angles are congruent since:
∠ ADC ≅ ∠ CBA [see here for definition of Congruent Triangles]

We can assume that ∠ DAB ≅ ∠ DCB since we could apply the same arguments #1 thru #6 to the diagonal DB as well.


Lemma 2: Parallelograms on the same base and in the same parallel are equal to each other.


(1) Let ABCD and EBCF be parallograms that share the same base BC and are colinear on AF.

(2) AD ≅ EF since:

AD ≅ BC [Since they are opposite sides of ABCD from Lemma 1 above]
EF ≅ BC [Since they are opposite sides of EBCF from Lemma 1 above]

(3) AE ≅ DF since:

AE = AD + DE
DF = EF + DE

AD ≅ EF (from the previous step)

(4) Now we can use Postulate 1 to conclude triangle ABE ≅ triangle DCF since:

AB ≅ DC [Since they are opposite sides of ABCD, from Lemma 1 above]

AE ≅ DF [Step #3]

∠ EAB ≅ ∠ FDC [since AB is parallel to DC and since Corresponding angles are congruent for parallel lines -- see here]

(5) We note that the trapezoid ABGD has the same area as EGCF since:

Both are formed from subtracting the area of DGE.

(6) But this implies that that the parallelogram ABCD is congruent to EBCF since they both formed by adding GBC to each trapezoid above.


Lemma 3: Triangles with equal bases in the same parallels are equal to each other.

If ABC, DEF are triangles with BC ≅ EF; if AD is parallel to BF; and if C,E lie on BF; then, ABC ≅ DEF.


(1) Let G be a point colinear with AD such that BG is parallel to AC.

(2) Let H be a point colinear with AD such that FH is parallel to DE.

(3) Then GACB and DHFE are parallelograms [Definition of parallelograms]

(4) Then the area of GACB is equal to the area of DHFE [See Lemma 2 above]

(5) The area of triangle ABC is half the area of GACB; and the area of triangle DEF is half the area of DHFE since:

(a) From Lemma 1 above, we know that each triangle such as DEF is congruent to its other half (in the case of DEF its other half is FHD)

(b) But if both triangles are congruent, then each triangle is (1/2) the total area, that is, the area of each triangle is half the area of each parallelogram.

(6) Since GACB ≅ DHFE (#4), we have (in terms of areas):
ABC = (1/2)GACB
DEF = (1/2)GACB

So we can see that ABC ≅ DEF.


Corollary 3.1: If a parallelogram has the same base with a triangle and is in the same parallels, then the parallelogram is double the triangle.


(1) Let ABCD be a parallelogram

(2) Triangle ABC ≅ triangle EBC from Lemma 3 above.

(3) And triangle ABC ≅ triangle CDA by Side-Angle-Side (see here) since:

(a) AD ≅ BC (By Lemma 1 above)

(b) DC ≅ AB (By Lemma 1 above)

(c) ∠ ADC ≅ ABC (By Lemma 1 above)

(4) Since triangle ABC ≅ triangle ECB ≅ triangle CDA is follows that parallelogram ABCD is double the area of triangle ECB.



Wednesday, March 22, 2006

Congruent Triangles

In today's blog, I review congruent triangles. These are all based on Euclid's Elements. If you would like to review Euclid's classic works, I strongly recommend David Joyce's web site on Euclid's Elements.

The site presents the complete Elements with proofs, definitions, and axioms and includes commentaries. All of the applets on this page are taken from David's web site.

Definition 1: Congruent Triangles

Two triangles are said to be congruent if corresponding angles are congruent and corresponding sides are congruent.

Postulate 1: Congruent by Side-Angle-Side (SAS)

For any two triangles, if two corresponding sides are congruent and the angle in between is congruent, then the triangles are congruent.

In the example below, if AB ≅ DE and BC ≅ EF and ∠ ABC ≅ ∠ DEF; then triangle ABC ≅ triangle DEF.

Euclid originally presented this as a proof using the method of superposition in his Elements (see here). The rigorousness of this method has been questioned. For these reasons, I am presenting this as a postulate. For those interested in more details, see here.

Postulate 2: Congruent by Side-Side-Side (SSS)

For any two triangles, if all 3 corresponding sides are congruent, then the triangles are congruent.

Euclid provides a proof of this one also using the method of superposition. For my purposes, I am presenting it as a postulate. For Euclid's proof, see here.

Lemma 1: For any two triangles, if two corresponding angles are congruent and the sides in between those angles are congruent, then the triangles are congruent.
If ∠ ABC ≅ ∠ DEF and BC ≅ EF and ∠ BCA ≅ EFD, then triangle ABC ≅ triangle DEF


(1) Assume that AB is not congruent to DE

(2) Since they are not congruent, either AB or DE is bigger. We will assume it is AB (if it were DE we could still use the same argument)

(3) Then there exists a point G on AB such that BG ≅ DE

(4) By our assumption, we have ∠ ABC ≅ ∠ DEF and we have BC ≅ EF

(5) So, then we have triangle GBC ≅ triangle DEF [By Postulate 1 above]

(6) But then ∠ BCG ≅ ∠ EFD since corresponding angles of congruent triangles are congruent.

(7) But this is impossible since ∠ BCA ≅ &EFD and ∠ BCA is clearly not equal to ∠ BCG.

(8) Since we have a contradiction we reject our assumption at (1) and conclude that AB ≅ DE

(9) But now we have triangle ABC ≅ triangle DEF by Postulate 1 above since:

AB ≅ DE [By Step #8]

∠ ABC ≅ ∠ DEF [By the given]

BC ≅ EF [By the given]


Corollary 1.1: For two triangles, if any two angles are congruent and any side is congruent, then both triangles are congruent.

(1) Let us assume that we have the pattern AAS (angle-angle-side) or SAA (side-angle-angle) congruent for the two triangles.

If we have ASA, then we know they are congruent by Lemma 1.

(2) But then we know that all corresponding angles of both triangles are congruent since:

Let a,b,c be the three angles of the first triangle. Let a',b',c' be the three angles for the second triangle.

Let's suppose that a,b match so that a' = a, b' = b.

Now, we know that a+b+c = 180 and a' + b' + c' = 180 [See Lemma 4 here]

So c = 180 - a - b

Also c' = 180 - a' - b' = 180 - a - b

So we see that c = c'

(3) And (#2) means that AAS and SAA imply ASA (angle-side-angle) which implies congruence by Lemma 1 above.