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THEOREM XLII.

252. A triangle is equivalent to half the rectangle of its base cid altitude.

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HYPOTHESIS.

A ABC, with base AC and altitude BD. CONCLUSION.

A ABC = half the rectangle of AC and BD. Proof. Through A draw AF || CB, and through B draw BF || CA, meeting AF in F;

A ACB ABFA.

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(217. The diagonal of a parallelogram bisects it.)

But D AFBC = rectangle of AC and BD,

(248. A parallelogram is equivalent to the rectangle of its base and altitude.)

A ABC = 1 o AFBC = 1 rectangle of AC and BD.

253. COROLLARY I. All triangles on the same base having their vertices in the same line parallel to the base, are equivalent.

254. COROLLARY II. Triangles having their vertices in the same point, and for their bases equal sects of the same line, are equivalent.

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

EXERCISES. 67. Equivalent triangles on equal bases have equal altitudes.

THEOREM XLIII.

256. If through any point on the diagonal of a parallelogram two lines be drawn parallel to the sides, the two parallelograms, one on each side of the diagonal, will be equivalent.

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HYPOTHESIS. P any point on diagonal BD of - ABCD; FG and HK lines through P|| AB and BC respectively, and meeting the four sides in the points F, G, H, K.

CONCLUSION. D AKPG = OPFCH.
PROOF. A ABD = A BCD.

AKBP = A BFP.
A GPD = A PHD.

(217. The diagonal of a parallelogram bisects it.)

From A ABD take away AKBP and A GPD, and we have left

AKPG. From the equal A BCD take away the equal As BFP and PHD, and we have left the o PFCH.

O AKPG

O PFCH.
(89. If equals be taken from equals, the remainders will be equal.)

257. In figures like the preceding, parallelograms like KBFP and GPHD are called parallelograms about the diagonal BD; while os AKPG and PFCH are called complements of parallelograms about the diagonal BD.

EXERCISES. 68. When are the complements of the parallelograms about a diagonal of a parallelogram congruent ?

V. Problems.

PROBLEM XII.

258. To describe a parallelogram equivalent to a given triangle, and having an angle equal to a given angle.

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GIVEN, A ABC and X G.

REQUIRED, to describe a parallelogram = A ABC, while having an angle = 4 G

CONSTRUCTION. Bisect AC in D.
At D, by 164, make . ADF = % G.
By 167, through A draw AH || DF.
By 167, through B draw BFH | CA.
AHFD will be the parallelogram required.
PROOF. Join DB. Then A ABD = ABCD,

(254. Triangles having their vertices in same point, and for bases equal sects of

the same line, are equivalent.)

A ABC is double A ABD.

But o AHFD is also double A ABD,

(255. If a parallelogram and a triangle be upon the same base and between the

same parallels, the parallelogram is double the triangle.)

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o AHFD = A ABC, and, by construction, ADF = 4 G.

PROBLEM XIII.

259. On a given sect as base, to describe a parallelogram equivalent to a given triangle, and having an angle equal to a given angle.

D K

F H

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GIVEN, the sect AB, A CDF, X G.

REQUIRED, to describe on AB a parallelogram = A CDF, while having an angle

CONSTRUCTION. By 258, make BHKL = A CDF, and having * HBL = * G, and place it so that BH is on line AB produced. Produce KL. Draw AM || BL. Join MB.

* HKM + * KMA = st. 4, (169. If a transversal cuts two parallels, it makes the two interior angles supple

mental.)

* HKM + % KMB < st. *. Therefore KH and MB meet if produced through H and B. (171. If a transversal cuts two lines, and the interior angles are not supplemental,

the lines meet.) Let them meet in Q. Through a draw QN || HA, and produce LB and MA to meet QN in P and N.

NPBA = OBHLK,
(256. Complements of parallelograms about the diagonal are equivalent.)

o NPBA = A CFD, and * ABP = X G.

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260. COROLLARY. Thus we can describe on a given base a rectangle equivalent to a given triangle.

PROBLEM XIV.

261. To describe a triangle equivalent to a given polygon.

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GIVEN, a polygon ABCDFG.
REQUIRED, to construct an equivalent triangle.

CONSTRUCTION. Join the ends of any pair of adjacent sides, as AB and BC, by the sect CA.

Through the intermediate vertex, B, draw a line || CA, meeting GA produced in H. Join HC.

Polygon ABCDFGA = polygon HCDFGH,
and we have obtained an equivalent polygon having fewer sides.

PROOF. A ABC = A AHC.
(253. Triangles having the same base and equal altitudes are equivalent.)
Add to each the polygon ACDFGA,

ABCDFGA HCDFGH. In the same way the number of sides may be still further diminished by one until reduced to three.

262. COROLLARY I. Hence we can describe on a given base a parallelogram equivalent to a given polygon, and having an angle equal to a given angle.

263. COROLLARY II. Thus we can describe on a given base a rectangle equivalent to a given polygon.

264. REMARK. To compare the surfaces of different polygons, we need only to construct rectangles equivalent to the given polygons, and all on the same base.

Then, by comparing the altitudes, we are enabled to judge of the surfaces.

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