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PROPOSITION XXXIX. THEOREM.

Equal triangles upon the same base, and upon the same side of it, are between the same parallels.

Let the equal As ABC, DBC be on the same base BC, and on the same side of it.

Join AD.

Then must AD be to BC.

For if not, through A draw AO || to BC, so as to meet BD, or BD produced, in O, and join OC.

But

Then As ABC, OBC are on the same base and between the same ||s,

:. Δ ABC= Δ OBC.

▲ ABC= ▲ DBC;

.. ▲ OBC= ▲ DBC,

the less the greater, which is impossible;

.. AO is not || to BC.

I. 37.

Hyp.

In the same way it may be shewn that no other line passing through A but AD is || to BC;

.. AD is to BC.

Q. E. D.

Ex. 1. AD is parallel to BC; AC, BD meet in E; BC is produced to P so that the triangle PEB is equal to the triangle ABC: shew that PD is parallel to AC.

D

Ex. 2. If of the four triangles into which the diagonals divide a quadrilateral, two opposite ones are equal, the quadrilateral has two opposite sides parallel.

8. E.

FROPOSITION XL. THEOREM.

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

E

D

Then ▲ ABC= ▲ OEF, between the same s.

But

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 || to BF.

For if not, through A draw AO || to BF, so as to meet ED, or ED produced, in O, and join OF.

they are on equal bases and I. 38.

Hyp.

A ABC= A DEF; ..▲ OEF=▲ DEF,

=

the less the greater, which is impossible.
.. AO is not || to BF.

In the same way it may be shewn that no other line passing through A but AD is || to BF,

.. AD is to BF.

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.

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

Then must

ABCD be double of ▲ EBC.

Join AC.

Then ▲ ABC= ▲ EBC, they are on the same base and between the same s ; I. 37.

and □ ABCD is double of ▲ ABC, :· AC is a diagonal of ABCD ; I. 34.

::ABCD is double of ▲ 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.

G

B

Let ABC be the given ▲, and D the given 4. It is required to describe a equal to ▲ ABC, having one of its s=LD.

Bisect BC in E and join AE.
At E make CEF= = LD.

Draw AFG || to BC, and from C draw CG | to EF
Then FECG is a parallelogram.

Now ▲ AEB▲ AEC,

they are on equal bases and between the same Is. .. AABC is double of ▲ AEC.

But FECG is double of ▲ AEC,

..they are on same base and between same [[s.
. FECG= ▲ ABC;

and FECG has one of its 48, CEF= D.
:: □ FECG has been described as was reqd.

I. 10.

I. 23.

I. 38.

I. 41.

Ax. 6.

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.

B

PROPOSITION XLIII.

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

THEOREM.

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

and AF, FC the others, which make up the whole figure ABCD,

and which are .. called the Complements.
Then must complement AF-complement FC.
BD is a diagonal of □ AC,
.. ▲ ABD= ▲ CDB ;

For

and. BF is a diagonal of HK,
.. ▲ HBF= ▲ KFB;

I. 34.

and. FD is a diagonal of EG, .. A EFD= ▲ GDF, Hence sum of ▲s HBF, EFD=sum of ▲s KFB, GDF. Take these equals from as ABD, CDB respectively, then remaining□ AF=remaining □ FC.

I. 34.

I. 34.

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 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.

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