But triangle A B C is equal to triangle D B C (a). by Axiom 1, (d) triangle EBC is equal to triangle DBC. The less equal to the greater, which is impossible. Therefore, the supposition (b) on which (d) is founded is impossible, and AE is not parallel to B C. In the same way it could be shown that every line drawn through A (except A D) is not parallel to B C. .. A D is parallel to B C. Q. E. D. I. EXERCISE. (Euc. I. 40.) Given. The triangles A B C, DEF equal to AA E one another; on equal bases B C, E F; in the same straight line BF, and on the same side of it. Required. To prove that ABC, DEF are between the same parallels. (Proceed, as in last theorem, by supposing the line joining A D not parallel to BF, draw another line meeting DE, and suppose that line parallel, and use Euc. I. 38.) II. EXERCISE. (Euclid I. 41) Given. The parallelogram ABCD and the triangle EBC, on the same base B C, and between the same parallels A E and B C. Required. To prove that A B CD is double of A B (Join A C. Then you have two triangles on the same base, and between the same parallels. Euc. I. 37 and I. 34.) Use D PROBLEM (Euclid I. 42). Repeat. The enunciations of Euc. I. 38, I. 41, and Axiom 6. Particular Enunciation. Given. The triangle ABC, and the angle D. Required. To describe a parallelogram equal to AB C, and having one of its an. gles equal to D. General Enunciation. To describe a parallelogram that shall be equal to a given triangle, and have one of its angles equal to a given angle. B Construction. (a) Bisect BC at E (by Euc. I. 10). Join AE (see next page). K C At the point E, in the straight line E C, (b) Make the angle C E F equal to angle D. (c) Through A draw A FG parallel to BC, and Proof. (d) Through C draw C G parallel to E F (by Euc. I. 31). By (c) and (d), and definition, The triangle ABE is equal to the triangle (e) .. The triangle ABC is double of the triangle A E Č. Again, the parallelogram FE CG and the triangle AEC are on the same base, and between the same parallels. .. by Euc. I. 41, (f) FECG is double of A E C. Hence, by (e) and (ƒ), and Axiom 6, its angle F E C is equal to angle D. Wherefore, a parallelogram FECG has been described equal to the given triangle A B C, and having one of its angles FEC equal to the given angle D. Q. E. F. EXERCISE. Given. The triangle AB C. another triangle, equal to (Produce AB, making BD equal to AB, and through C draw CE parallel to AD. Describe a triangle on BD having its vertex in CE; and use Euc. I. 38 for proof.) THEOREM (Euclid I. 43). Repeat. The enunciation of Euc. I. 34, and Axioms 2 and 3. General Enunciation. The complements of the parallelograms which are about the diameter of a parallelogram, are equal to one another. EXPLANATION. The parallelograms AEKH, H A E K D F B G KG CF are said to be described about the diameter A C. And the remaining parts of the whole parallelogram —that is, the parallelogram E B GK, and the parallelogram HKFD-are called the complements of the parallelograms about the diameter. Particular Enunciation. Given. The parallelogram A B CD, having the parallelograms AEKH, KGCF about the diameter A C. Required. To prove that E B G K is equal to HKFD. Proof. ABCD is a parallelogram, and A C its diameter. .. by Euc. I. 34, (a) the triangle ABC is equal to the triangle ADC. AEKH is a parallelogram, and A K its diameter. by Euc. I. 34, (b) the triangle A E K is equal to the triangle A H K. KGCF is a parallelogram, and KC its diameter. by Euc. I. 34, (c) the triangle KG C is equal to the triangle KFC. Take away the triangles AEK, KG C, from the triangle ABC, the remainder is the parallelogram E B G K. Take away the triangles AHK, KFC from the triangle ADC, the remainder is the parallelogram HKFD. Hence, by (a), (b), and (c), and Axiom 6, Q. E. D. |