EXERCISE. Given. The triangle AB C. another triangle, equal to (Produce AB, making BD equal to AB, and through C draw CE parallel to A D. 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, E A E K B KGCF are said to be described about the diameter A C. And the remaining parts of the whole parallelogram -that is, the parallelogram EB 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, KG CF, 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 A D C. 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. KG CF is a parallelogram, and KC its diameter. .. by Euc. I. 34, (c) the triangle KGC 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. PROBLEM (Euclid I. 44). Repeat. The enunciations of Euc. I. 29, I. 41, and Axioms 1 and 12. General Enunciation. To a given straight line to apply a parallelogram, which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle. Particular Enunciation. line AB, the triangle Construction. E D G B Make the parallelogram BEFG equal to the triangle C, and having the angle EBG equal to H M K D L the angle D, so that BE may be in the same straight line with AB (by Euc. I. 42). Produce FG to H. Through A draw A H parallel to B G and EF (by Euc. I. 31). Join HB. Then AH is parallel to EF, and HF falls upon them. ... AHF, HFE are together equal to two right angles. .. BHF, HFE are together less than two right angles. And so, by Axiom 12, HB and FE will meet if produced. Let them be produced and meet at K. Produce HA, GB, to L and M. HK is the diameter of the parallelogram FHLK. LB and B F are the complements of the parallelograms about HK. ... by Euc. I. 43, LB is equal to B F. But BF was made equal to C. .. by Axiom 1, LB is equal to C. By Construction, angle CBE is equal to angle D. By Euc. I. 15, angle GBE is equal to angle ABM. .. by Axiom 1, angle A B M is equal to angle D. Wherefore to the given straight line A B, the parallelogram LB is applied equal to the triangle C, and having the angle A B M equal to the angle D. Q. E. F. PROBLEM (Euclid I. 45). Repeat. The enunciations of Euc. I. 14, 29, 30, and Axioms 2 and 2a. General Enunciation. To describe a parallelogram equal to a given rectilineal figure, and having an angle equal to a given rectilineal angle. Particular Enunciation. Given. The rectilineal figure A B C D, and the angle E. D B C Required. To describe a parallelogram equal to ABCD, with an angle equal to E. Construction. Join BD. Describe the parallelogram FH equal to the triangle ADB, and having the angle FGH equal to the angle E (by Euc. I. 42). Ἐ G K To the straight line GH apply the parallelogram KM, equal to the triangle DBC, and having the angle KHM equal to the angle E (by Euc. I. 44). The figure FGML is the required parallelogram. H M |