PROP. XLI. THEOREM. If a parallelogram and a triangle be on the same base, and between the same parallels, the parallelogram shall be double of the triangle. Let ABCD be a parallelogram, and EBC a triangle, on the same base BC, and between the same parallels BC and AE. Then it is to be proved that The parallelogram ABCD is double the triangle EBC. PROOF.-Because the triangles ABC and EBC are upon the same base BC, and between the same parallels BC and AE, therefore the triangle ABC the triangle EBC (I. 37). = But the parallelogram ABCD is double the triangle ABC (L. 34). Therefore, it is proved, as required, that The parallelogram ABCD is double the triangle EBC. Wherefore, If a parallelogram and a triangle, &c. Exercise. Q. E. D. If ABCD be a quadrilateral with base AB and diagonals AC and BD, bisecting each other in E; prove that ABCD is a parallelogram. PROP. XLII. PROBLEM. To describe a parallelogram that shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle. Let ABC be the given triangle, and D the given rectilineal angle. It is required to describe a parallelogram=the triangle CONSTRUCTION.-1. Bisect BC in E (I. 10), and join AE. 2. At the point E in the line EC, make the angle CEF= the angle D (I. 23). 3. Through A draw AG parallel to EC, and through C draw CG parallel to EF (I. 31). Then it is to be proved that FECG is a parallelogram the triangle ABC, and the angle D. having the angle CEF PROOF.-Because the triangles ABE and AEC are upon equal bases BE and EC (cons.), and between the same parallels BC and AG (cons.), therefore the triangle ABE = the triangle AEC (I. 38), and therefore the triangle ABC is double the triangle AEC. But because the parallelogram FECG (cons. and def. 34) and the triangle AEC are upon the same base EC, and between the same parallels EC and AG, therefore the parallelogram FECG is double the triangle AEC (I. 41). = But we have seen that the triangle ABC is double the triangle AEC, therefore the parallelogram FECG the triangle ABC (ax. 6), and it has one of its angles CEF = the angle D (cons.). the triangle ABC, and Therefore, it is proved, as required, that Q. E. F. The complements of the parallelograms which are about the diameter of any parallelogram are equal to one another. Let ABCD be a parallelogram, of which AC is a diameter, with the parallelograms AEKH and KGCF about the diameter, i.e. through which the diameter AC passes. Then the parallelograms EBGK and HKFD, which make up the whole parallelogram ABCD, are the complements, and it is to be proved that The complement EBGK = the complement HKFD. PROOF.-Because ABCD is a parallelogram, and AC its diameter, therefore the triangle ABC the triangle ADC (I. 34). Similarly the triangle AEK = the triangle AHK, and the triangle KGC = the triangle KFC. Hence the triangles AEK and KGC taken together = the triangles AHK and KFC taken together (ax. 2). But we have seen that the whole triangle ABC = the whole triangle ADC; therefore the remainder, or the parallelogram EBGK = the remainder, or the parallelogram HKDF (ax. 3); and these are the complements in the parallelogram ABCD. Therefore, it is proved, as required, that The complement EBGK = the complement HKFD. Wherefore, The complements of the parallelograms, &c. Q. E. D. PROP. XLIV. PROBLEM. 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. Let AB be the given straight line; C the given triangle; and D the given rectilineal angle. It is required to apply to the straight line AB a parallelogram the triangle C, and containing an angle the angle D. CONSTRUCTION (1).—1. Make the parallelogram BEFG = the triangle C, and with an angle EBG= the angle D (I. 42), so that BE may be in the same right line with AB. 2. Produce FG to H. 3. Through A draw AH parallel to BG or EF (I. 31), and join HB. = PROOF (1).-Because the straight line HF falls on the parallel lines AH and EF, therefore the two interior angles AHF and HFE two right angles (I. 29), and therefore the angles BHF and HFE are less than two right angles; consequently the straight lines HB and FE will meet, if produced far enough (ax. 12). CONSTRUCTION (2).-1. Produce the straight lines HB and FE to meet as in K. 2. Through K draw KL parallel to EA or FH (I. 31), and produce HA and GB to the points L and M. Then it is to be proved that ABML is a parallelogram the triangle C, applied to the straight line AB, and having the angle ABM= the angle D. PROOF (2). Because FL* is a parallelogram (cons. and def. 34) of which HK is a diameter, and FB and BL complements of the parallelograms GA and EM, about the diameter, therefore the parallelogram FB the parallelogram BL (I. 43). But the parallelogram FB the triangle C (cons.), therefore the parallelogram BL the triangle C (ax. 1), and it is applied to the line AB, for it is one of its sides. Again, because the angle GBE = the angle D (cons.), and because the angle ABM = the angle GBE (I. 15), therefore the angle ABM = the angle D (ax. 1). Therefore, it is proved, as required, that ABML is a parallelogram the triangle C, applied to the straight line AB, and having the angle ABM = the angle D. Q. E. F. * N.B.-Parallelograms may be referred to by the two letters standing at opposite angles, instead of the four, at each angle. Thus the parallelogram FHLK may be referred to as the parallelogram FL, or KH. Exercises. 1. If ABCD be a parallelogram with diagonal AC= diagonal BD; prove that each angle of the parallelogram A, B, C, and D is a right angle. 2. If ABCD be a parallelogram with its diagonals AC and BD intersecting at right angles in E; prove that the sides of the parallelogram AB, BC, CD, and DA = each other. |