The parallels (EF, GH) through any point (K) in one of the diagonals (AC) of a parallelogram divide it into four parallelograms, of which the two (BK, KD) through which the diagonal does not pass, and which are called the complements of the other two, are equal. Dem.—Because the diagonal bisects the parallelograms AC, AK, KC, we have [xxxiv.] the triangle ADC equal to the triangle ABC, the triangle AHK equal to AEK, and the triangle KFC equal to the triangle KGC. Hence, subtracting the sums of the two last equalities from the first, we get the parallelogram DK equal to the parallelogram KB. Cor. 1.—If through a point K within a parallelogram ABCD lines drawn parallel to the sides make the parallelograms DK, KB equal, K is a point in the diagonal AC. Cor. 2.—The parallelogram BH is equal to AF, and BF to HC. Cor. 2. supplies an easy demonstration of a fundamental Proposition in Statics. Exercises. 1. If EF, GH be parallels to the adjacent sides of a parallelogram ABCD, the diagonals EH, GF of two of the four s into which they divide it and one of the diagonals of ABCD are concurrent. Dem.—Let EH, GF meet in M; through M draw MP, MJ parallel to AB, BC. Produce AD, GH, BC to meet MP, and AB, EF, DC to meet MJ. Now the complement OF = FJ: to each add the FL, and we get the figure OFL = CJ. Again, the complement PH = HK [xliii.]: to each add the OC, and we get the PC = figure OFL. Hence the PC = CJ. Therefore they are about the same diagonal [xliii., Cor. 1]. Hence AC produced will pass through M. 2. The middle points of the three diagonals AC, BD, EF of a quadrilateral ABCD are collinear. Dem.—Complete the AEBG. Draw DH, CI parallel to AG, BG. Join IH, and produce; then AB, CD, IH are concurrent (Ex. 1); therefore IH will pass through F. Join EI, EH. Now [xi., Ex. 2, 3] the middle points of EI, EH, EF are collinear, but [xxxiv., Ex. 1] the middle points of EI, EH are the middle points of AC, BD. Hence the middle points of AC, BD, EF are collinear. PROP. XLIV.—Problem. To a given, right line (AB) to apply a parallelogram which shall be equal to a given triangle (C), and have one of its angles equal to a given angle (D). Sol.—Construct the parallelogram BEFG [xlii.] equal to the given triangle C, and having the angle B equal to the given angle D, and so that its side BE shall be in the same right line with AB. Through A draw AH parallel to BG [xxxi.], and produce FG to meet it in H. Join HB. Then because HA and FE are parallels, and HF intersects them, the sum of the angles AHF, HFE is two right angles [xxix.]; therefore the sum of the angles BHF, HFE is less than two right angles; and therefore (Axiom xii.) the lines HB, FE, if produced, will meet as at K. Through K draw KL parallel to AB [xxxi.], and produce HA and GB to meet it in the points L and M. Then AM is a parallelogram fulfilling the required conditions.