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.
To make a parallelogram

= 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.


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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
in the diagonal AC let
EG be drawn || AD (1.31)
and FH || AB, then FE
and GH are the parallelo-
grams about the diagonal,
and FG and KB are called the complements of the
parallelograms about the diagonal, because with

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.


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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.



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Given A, 2B, and st. line CD.
To apply to CD a parallelogram (i.e., so that CD shall

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


To describe a parallelogram equal to a given recti. lineal figure, and having an angle equal to a given angle.



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Given à rectilineal figure ABCD and _ E.
To construct a parallelogram equal to ABCD, and

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.

2 LI:S.;


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To describe a square on a given straight line.
Given st. line AB.
To construct a square on AB.
Draw AC I AB (1. 11), and make

AC = AB (Post. 3).
Through C draw CD || AB, and
through B draw BD || AC (1.31);

:: 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.

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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.

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