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PROP. 42. PROB.
To describe a parallelogram that shall be equal to a given triangle, and have one of its angles equal to a given angle.
Given A ABC and 2 D.
= A ABC, and having
an angle = LD. Bisect BC at E (I. 10); at
E in EC make CEF
= LD (I. 23). Through A draw AFG || BC (1. 31), and through C
draw CG || EF. FECG is the parallelogram required. Join AE. :: BE = EC (cons.), A ABE = AAEC (I. 38),
A ABC is double of A AEC. But FC is double of A AEC (1. 41). :: | FC = AABC and has 2 FEC = · D.
Note. This proposition is the first step of a set of four (I. 42, 44, 45, and II. 14), which show that it is possible to find a square equal to any rectilineal figure. The whole problem is called the “Quadrature of a Rectilineal Area."
127. Construct a triangle equal to a given parallelogram, and having an angle equal to a given angle.
128. Construct a triangle equal in area to a given triangle, whose base shall be any part of the base of the given triangle.
129. Construct a triangle equal in area to a given triangle, whose vertex shall be a given pt. in one of the sides of the given triangle.
130. Construct a triangle equal to a given pentagonal figure, whose base shall be in the same straight line with a side of the pentagon, and whose vertex shall be the opposite angular point of the pentagon.
131. To construct a polygon of n-1 sides, equal in area to a polygon of n sides, ñ being greater than three.
PROP. 43. THEOR. The complements of the parallelograms which are about the diagonal of any parallelogram are equal to one another. Given parallelogram ABCD, A
and through K any point
these they make up the whole parallelogram. To prove FG KB. By I. 34, A ADC A ABC, and parts of these, for a
like reason, are equal to one another, viz., A AFK = A AEK, and AKGC : AKHC.
:: remainder FG=remainder KB(Ax.3). Q.E.D. 132. The “complements” are equiangular to one another and to the whole parallelogram.
133. The “parallelograms about the diagonal” of a square are squares.
134. If through a point K, within a parallelogram, lines FKH, EKG be drawn parallel to the sides, and making KD
KB, K shall be on the diagonal AC.
135. To construct a quadrilateral equal in area to a given quadrilateral, on one side of the given quadrilateral as base, and having for the opposite side a line parallel to this drawn through a given point in the corresponding side of the given quadrilateral.
136, To describe a parallelogram which, both in perimeter and in area, shall be equal to a given triangle.
137. Any straight line drawn through the middle point of the diagonal of a parallelogram to meet the sides bisects the parallelogram.
PROP. 44. PROB. 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 angle.
Given A, 2B, and st. line CD.
be one of its sides) equal to AA and having one
of its angles equal to 2 B. Make a parallelogram CEFG equal to A A having
4 FGC = LB (I. 42), and place it with one side in the same straight line with CD. Produce FG. Through D draw DH || CG (1. 31). Join HC. : FH meets the parallels FE and HD, :: LEFH +
LFHD 2 l's. (I. 29). :: _ EFH + _ FHC are < 2 Ls., :. FE and HC pro
duced will meet towards EC (1. 29, Cor.). Let them meet at K. Through K draw KLM || FH
(1. 31), and produce GC and HD to L and M. :: FM is a parallelogram, and HK its diagonal. :. FC = CM (1. 43); but FC=A (cons.), :: CM = A. Again, :: _ CLM = _ FGC (1. 29) = _ B (cons.),
I. _ : CLM
AA, has – CLM _ B, and is applied to the given line CD. Q.E.F.
LB. :: 0 CM
PROP. 45. PROB.
To describe a parallelogram equal to a given recti. lineal figure, and having an angle equal to a given angle.
Given à rectilineal figure ABCD and _ E.
having an angle equal to LE. Divide the given figure into triangles I., II. Make
IFK = A I., with 2 FGH = _E (1. 42). To KH apply O KM = A II., with 2 KHM = _E(1.44).
= : - KHM - E, and 2 FGH LE (cons.), :: LFGH
- KHM. To each add _ GHK; :: LFGH + LGHK < GHK + L KHM. But _FGH + L GHK = 2 L"s. (1. 29), :: L GHK + < KHM and :: GH and HM are in the same st. line (1. 14). Again, :: FK is || GH, and KL is || HM, :: FK and KL are both || GM, :: FK and KL are in same st. line (Ax. 11). :: FG is | KH and KH is || LM, : FG is | LM (1.30), .: FM is a parallelogram. and :: FH = A I., and KM ΔΙI., : 0 FM AC, and has LG
Q.E.F. Note.-As the given figure may have any number of sides it may give more triangles than two. Then to LM apply a parallelogram equal to the third triangle with L at M=_ E, and so on as above.
To describe a square on a given straight line.
AC = AB (Post. 3).
:: ACDB is a parallelogram, and :: AB=AC (cons.), all the sides are equal (1. 34). Again, :: _ BAC + ACD 2 Ls. (I. 29), and
BAC is a rt. angle (cons.), :: LACD is a rt. angle, and .:: all the angles are right
angles (1. 34), :: ABDC is a square on the given line AB. Q.E.F.
Note-We here construct a parallelogram with one angle a right angle, and two adjacent sides equal, and then prove that it is a square.
138. Construct a rectangle whose sides shall be respectively equal to two given straight lines.
139. Squares on equal lines are equal; and, conversely, equal squares are on equal lines.
140. Of all parallelograms of equal area the square has the least perimeter.
141. To bisect a square by a straight line drawn through a given point in one of its sides ; (2) to divide a square into four equal parts by lines drawn through a given point in one of its sides.
142. If a straight line be drawn bisecting the two nonparallel sides of a trapezium, it will be parallel to the other sides and equal to half their sum. Also the area of the trapezium is equal to that of a rectangle contained by this line and the perpendicular between the parallel sides of the trapezium.