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.
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):
BC ≅ AD
AB ≅ DC
(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.