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Equal triangles upon equal bases, in the same straight line, and towards the same parts, are between the same parallels.

D

E

Let the equal As ABC, DEF be on equal bases BC, EF in the same st. line BF and towards the same parts.

Join AD.

Then must AD be Il to BF.

For if not, through A draw A 0 || to BF, so as

meet ED, or ED produced, in 0, and join OF.

Then a ABC= A OEF, :: they are on equal bases and between the same | s.

I. 38. But A ABC= A DEF;

Нур. .: A OEF= A DEF, the less=the greater, which is impossible.

.. AO is not ll to BF. In the same way

it
may

be shewn that no other line passing through A but AD is || to BF,

.. AD is ll to BH.

Q. E. D.

Ex. 1. If the triangles be not towards the same parts, shew that the straight line joining the vertices of the triangles is bisected by the line containing the bases.

Ex. 2. The straight line, joining the points of bisection of two sides of a triangle, is parallel to the base.

Ex. 3. The straight lines, joining the middle points of the sides of a triangle, divide it into four equal triangles.

PROPOSITION XLI. THEOREM.

If a parallelogram and a triangle be upon the same base, and between the same parallels, the parallelogram is double of the triangle.

Let the O ABCD and the A EBC be on the same base BC and between the same ||s AE, BC.

Then must D ABCD be double of - EBC.

Join AC.

Then ABC= A EBC, :: they are on the same base and between the same lls ;

I. 37. and D ABCD is double of A ABC, :: AC is a diagonal of ABCD;

I. 34. ..O ABCD is double of A EBC.

Q E. D.

Ex. 1. If from a point, without a parallelogram, there be drawn two straight lines to the extremities of the two opposite sides, between which, when produced, the point does not lie, the difference of the triangles thus formed is equal to half the parallelogram.

Ex. 2. The two triangles, formed by drawing straight lines from any point within a parallelogram to the extremities of its opposite sides, are together half of the parallelogram.

PROPOSITION XLII. PROBLEM. To describe a parallelogram that shall be equal to a given triangle, and have one of its angles equal to a given angle.

A F

B

Let ABC be the given A, and D the given b. It is required to describe a o equal to A ABC, having one of its 68=.D. Bisect BC in E and join AE.

1. 10. At E make CEF= .D.

I. 23. Draw AFG || to BC, and from C draw CG 1| to EF.

Then FECG is a parallelogram. Now A AEB= A AEC, :: they are on equal bases and between the same lls.

I. 38. ..A ABC is double of A AEC. But O FECG is double of A AEC, :: they are on same base and between same is.

I. 41. :O FECG= A ABC;

Ax. 6. and O FECG has one of its 2 s, CEF= _ D. .::O FECG has been described as was reqd.

Q. E. F.

Ex. 1. Describe a triangle, which shall be equal to a given parallelogram, and have one of its angles equal to a given rectilineal angle.

Ex. 2. Construct a parallelogram, equal to a given triangle, and such that the sum of its sides shall be equal to the sum of the sides of the triangle.

Ex. 3. The perimeter of an isosceles triangle is greater than the perimeter of a rectangle, which is of the same altitude with, and equal to, the given triangle.

PROPOSITION XLIII. THEOREM. The complements of the parallelograms, which are about the diameter of any parallelogram, are equal to one another.

H

Let ABCD be a O, of which BD is a diagonal, and EG, HK the Os about BD, that is, through which BD passes,

and 'AF, FC the other Os, which make up the whole figure ABCD,

and which are .. called the Complements.
Then must complement AF=complement FC.
For :: BD is a diagonal of D AC,
A ABD= A CDB;

I. 34.
and :: BF is a diagonal of O HK,
.:. Δ HBF = ΔKFB;

I. 34. and :: FD is a diagonal of O EG, . AEFD= AGDF.

I. 34. Hence sum of As HBF, EFD=sum of AS KFB, GDF. Take these equals from As ABD, CDB respectively, then remaining O AF=remaining O FC. Ax. 3.

Q. E. D. Ex. 1. If through a point 0, within a parallelogram ABCD, two straight lines are drawn parallel to the sides, and the parallelograms OB, OD are equal ; the point O is in the diagonal AC.

Ex. 2. ABCD is a parallelogram, AMN a straight line meeting the sides BC CD (one of them being produced) in M, N. Shew that the triangle MBN is equal to the triangle MDC.

PROPOSITION 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 angle.

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Let AB be the given st. line, C the given A, D the given ..

It is required to apply to AB a O=A C and having one of its 28= 2D.

Make a O=A C, and having one of its angles= LD, I. 42. and suppose it to be removed to such a position that one of the sides containing this angle is in the same st. line with AB, and let the O be denoted by BEFG.

Produce FG to H, draw AH || to BG or EF, and join BH. Then :: FH meets the ||s AH, EF, .: sum of 28 AHF, HFE=two rt. 28;

I. 29. .: sum of 2s BHG, HFE is less than two rt. 28; .: HB, FE will meet if produced towards B, E. Post. 6.

Let them meet in K.
Through K draw KL || to EA or FH,
and produce HA, GB to meet KL in the pts. L, M.
Then HFKL is a 0, and HK is its diagonal ;

and AG, ME are Os about HK,
.. complement BL=complement BF, I. 43.

..O BL= AC. Also the O BL has one of its ZS, ABM= 2 EBG, and .. equal to 2 D.

Q. E. F.

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