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Join PA, and from the angle P bisect thetrapezium APCD (iv.19.) by the line PE. On PE make the triangle PEF equal to ABP. Bisect EF in G; join PG. PG bisects the trapezium.

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P

Since FG is equal to GE, the triangle PGF is equal to the triangle PGE. But PGE is equal to half the triangle ABP; and PEC is half the figure PABC; whence PGC is half of the trapezium ABCD; which is .. bisected by PG.

(21.) If two sides of a trapezium be parallel; the triangle contained by either of the other sides, and the two straight lines drawn from its extremities to the bisection of the opposite side, is half the trapezium.

Let ABCD be a trapezium, having the side AB parallel to DC. Let AD be bisected in E; join BE, CE; the triangle BEC is half of the trapezium.

F

E

B

Through E draw FEG parallel to BC, meeting CD in G, and BA produced in F. The alternate angles FAE, EDG being equal, as also the angles at E, and AE-ED,.. the triangles AEF, DEG are equal; whence the parallelogram BFGC is equal to the trapezium ABCD. But BFGC and the triangle BEC, being on the same base BC, and between the same parallels BC, FG, the triangle BEC is half of BFGC, and .. also half of ABCD.

COR. From the demonstration it appears, that a trapezium which has two sides parallel, may be reduced to a parallelogram equal to it, by drawing through the point of bisection of one of the sides, which are not parallel, a line parallel to the other of those sides, and meeting the parallel sides.

(22.) To divide a given trapezium whose opposite sides are parallel, in a given ratio, by a line drawn through a given point, and terminated by the two parallel sides.

Let ABCD be a trapezium whose sides AB, DC are parallel, and P the given point. Bisect AD in E, and draw EF parallel to AB, meeting BC in F. Divide EF in

P

G in the given ratio; and through G and P draw IPH; IPH will divide the trapezium in the given ratio.

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Draw KGL, MFN parallel to AD; then EA=KG MF, and ED=GL=FN; but AE=ED, ;. KG= GL, and MF=FN; whence (iv. 21. Cor.) ADIH= AL, and HICB = KN.

Now AL KN:: EG: GF, .. ADIH : HICB :: EG : GF,

i. e. in the given ratio.

(23.) If a trapezium, which has two of its adjacent angles right angles, be bisected by a line drawn from the middle of one of those sides which are not parallel; the

sum of the parallel sides will have to one of them the same ratio, that the side which is not bisected has to that segment of it which is adjacent to the other.

Let ABCD be a trapezium, having the angles at A and D right angles, and ... AB, DC parallel; and let the trapezium be bisected by EF; if AD be bisected in E, AB+DC : AB :: BC ; CF;

but if BC be bisected in F,

AB+ DC: AB :: AD : DE.

H

Produce DA, CB to meet in G. Join AF, DF, BE, CE, and let fall the perpendiculars AH, DI. Since AE-ED, the triangles AFE, DFE are equal, .. the triangles DFC, AFB are equal; .. the rectangles FG, DI and BF, AH are equal,

and FC: FB :: AH : DI :: AB : CD, :. AB+CD : AB :: (FC+FB=) BC : FC. But if BC be bisected, the triangles A EBF, ECF being equal, the triangles AEB, EDC are also equal,

.. (Eucl. vi. 15.) AB : CD :: DE : EA,

E

D

and AB+ CD : AB :: (DE+EA=) AD : De. In like manner AB + DC: DC :: AD : AE.

F

(24.) If the sides of an equilateral and equiangular pentagon be produced to meet; the angles formed by these lines are together equal to two right angles.

Let ABCDE be an equilateral and equiangular

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pentagon; and let the sides be duced to meet in F, G, H, I, K; the angles at these points are together equal to two right angles.

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D

B

C

For since BCG is the exterior angle of the triangle FCI, it is equal to the angles at F and I. For the same reason the angle CBG is equal to the angles at K and H; and the angles at F, G, H, I, K are equal to the three angles of the triangle BCG, i. e. to two right angles.

(25.) If the sides of an equilateral and equiangular hexagon be produced to meet; the angles formed by these lines are together equal to four right angles.

Let ABCDEF be an equilateral

and equiangular hexagon; and let the sides be produced to meet in G, H, I, K, L, M ; the angles at these points are together equal to four right angles.

For GLI being a triangle, the angles at G, I, L are equal to two

M

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right angles; and for the same reason, the angles at H, K, M are equal to two right angles; .. the six angles are equal to four right angles.

(26.) The area of any two parallelograms described on the two sides of a triangle is equal to that of a paral

lelogram on the base, whose side is equal and parallel to the line drawn from the vertex of the triangle to the intersection of the two sides of the former parallelograms produced to meet.

Let BE and CG be parallelograms described on the sides AB, BC of the triangle ABC; and let EF, HG be produced to meet in D. Join DB; produce it, and make IK=DB; through A draw AL equal and parallel to IK; and complete the parallelo

E

D

gram AM. AM is equal to AF and CG together.

H

Produce LA, MC to N and O; since ND is parallel to AB, and AN to BD, NABD is a parallelogram, and equal to EB, which is on the same base, and between the same parallels. It is also equal to AK; because they are upon equal bases DB, IK, and between the same parallels; .. AK=EB. In the same manner IM=BH, .. AM is equal to AF and CG together.

(27.) The perimeter of an isosceles triangle is greater than the perimeter of a rectangular parallelogram, which is of the same altitude with, and equal to the given triangle.

Let ABC be an isosceles triangle, whose base is BC. Draw AE perpendicular to BC, and... bisecting it; and draw AD, CD parallel respectively to BC,

B

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