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base and the adjacent one as the altitude. The rectangle having two given segments as its base and altitude is called the rectangle on these segments.

Notation.The symbol stands for the word rectangle and o for parallelogram.

Rectangles and parallelograms are commonly indicated by naming a pair of their opposite vertices.

SECTION 1.

COMPARISON OF AREAS--RECTANGLES,

PARALLELOGRAMS, TRIANGLES.

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139o. Theorem.1. Rectangles with equal bases and equal altitudes are equal.

2. Equal rectangles with equal bases have equal altitudes. 3. Equal rectangles with equal altitudes have equal bases. 1. In the Os BD and FH, if

AD=EH, and

AB=EF, then

DBD=DFH. Proof.--Place E at A and EH along AD. Then, as ZFEH=LBAD= 7, EF will lie along AB.

And because EH=AD and EF=AB, therefore H falls at D and F at B, and the two os are congruent and therefore equal.

2. If OBD=OFH and AD=EH, then AB=EF.

Proof.—If EF is not equal to AB, let AB be > EF.
Make AP=EF and complete the OPD.
Then

OPD=OFH, by the first part,
but
OBD=OFH,

(hyp.)

q.e.d. OPD=OBD, which is not true, ::: AB and EF cannot be unequal, or AB=EF.

9.e.d. 3. If OBD=DFH and AB=EF, then AD=EH.

Proof.Let AB and EF be taken as bases and AD and EH as altitudes (138°), and the theorem follows from the second part.

9.e.d. Cor. In any rectangle we have the three parts, base, altitude, and area. If any two of these are given the third is given also.

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140°. Theorem.-A parallelogram is equal to the rectangle

ç on its base and altitụde.

AC is a O whereof AD is the base and DF is the altitude.

Then OAC=O on AD and DF. Proof.Complete the DADFE by drawing AE I to CB produced.

Then AAEB= ADFC, ::: AE=DF, AB=DC, and

LEAB=LFDC; ::: ADFC may be transferred to the position AEB, and OABCD becomes the DAEFD, JAC=O on AD and DF.

q.e.d. Cor. 1. Parallelograms with equal bases and equal altitudes are equal. For they are equal to the same rectangle.

Cor. 2. Equal parallelograms with equal bases have equal altitudes, and equal parallelograms with equal altitudes have equal bases.

Cor. 3. If equal parallelograms be upon the same side of the same base, their sides opposite the common base are in line.

141°. Theorem.-A triangle is equal to one-half the rectangle on its base and altitude.

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E

ABC is a triangle of which AC is the base and BE the altitude.

Then AABC=žo on AC and BE.

Proof.Complete the OABDC, of which AB and AC are adjacent sides. Á Then

AABC = ADCB, .. AABC=iOAD={o on AC and BE. (140°) q.e.d.

Cor. 1. A triangle is equal to one-half the parallelogram having the same base and altitude.

Cor. 2. Triangles with equal bases and equal altitudes are equal. For they are equal to one-half of the same rectangle.

Cor. 3. A median of a triangle bisects the area. For the median bisects the base.

Cor. 4. Equal triangles with equal bases have equal altitudes, and equal triangles with equal altitudes have equal bases.

Cor. 5. If equal triangles be upon the same side of the same base, the line through their vertices is parallel to their common base.

B

142°. Theorem.-If two triangles are upon opposite sides of the same base

1. When the triangles are equal, the base bisects the seg. ment joining their vertices;

2. When the base bisects the segment joining their vertices, the triangles are equal. (Converse of 1.)

ABC and ADC are two triangles upon opposite sides of the common base AC.

\H

H F
1. If AABC=AADC,
then

BH=HD,
Proof.Let BE and DF be altitudes,
Then ::

AABC=AADC, .. BE=DF,

AEBH=AFDH, and BH=HD. q.e.d. 2. If BH=HD, then AABC=AADC.

A

E

Proof.Since BH=HD, ... ΔΑΒΗ =ΔΑΡΗ, and ACBH=ACDH.

(141°, Cor. 3) .. adding, AABC=AADC.

q.e.d.

143. Def.-By the sum or difference of two closed figures is meant the sum or difference of the areas of the figures.

If a rectangle be equal to the sum of two other rectangles its area may be so superimposed upon the others as to cover both.

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144o. Theorem.-If two rectangles have equal altitudes, their sum is equal to the rectangle on their common altitude and the sum of their bases.

Proof.—Let the Os X and Y, having equal altitudes, be so placed as to have

E their altitudes in common at CD, and so that one may not overlap the other. Then

LBDC=LCDF= 7,
BDF is a line.

(38°, Cor. 2) Similarly

ACE is a line. But BD is || to AC, and BA is || to DC || to FE; therefore AF is the o on the altitude AB and the sum of the bases AC and CE; and the DAF=DAD +OCF.

q.e.d. Cor. 1. If two triangles have equal altitudes, their sum is equal to the triangle having the same altitude and having a base equal to the sum of the bases of the two triangles.

Cor. 2. If two triangles have equal altitudes, their sum is equal to one-half the rectangle on their common altitude and the sum of their bases.

Cor. 3. If any number of triangles have equal altitudes, their sum is equal to one-half the rectangle on their common altitude and the sum of their bases.

In any of the above, “base” and “altitude” are interchangeable.

145o. Theorem.Two lines parallel to the sides of a parallelogram and intersecting upon a diagonal divide the parallelogram into four parallelograms such that the two through which the diagonal does not pass are equal to one another.

In the ZABCD, EF is || to AD and GH is || to BA, and these intersect at O on the diagonal AC.

Then 7 BO=OOD.

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A

HD

Proof.-AABC=AADC, and AAEO=AAHO, and AOGC=AOFC;

(141°, Cor. 1) but OBO=AABC-AAEO-AOGC, and ZOD=AADC-AAHO-A0FC. BO=OOD.

q.e.d.

[blocks in formation]

Cor. 2. If OBO=DOD, O is on the diagonal AC. (Converse of the theorem.)

For if () is not on the diagonal, let the diagonal cut EF in O'. Then BO'=OO'D.

(145) But OBO' is < OBO, and OO'D is > DOD; .. BO is > DOD, which is contrary to the hypothesis; .. the diagonal cuts EF in 0.

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Ex. Let ABCD be a trapezoid. (84°, Def.) In line with AD make DE=BC, and in line with BC make CF=AD.

Then BF=AE and BFEA is a J.

But the trapezoid CE can be superimposed on the trapezoid DB, since the sides are respectively equal, and

ZF=A, and ZE=B, etc.

trapezoid BD="OBE, or, a trapezoid is equal to one-half the rectangle on its altitude and the sum of its bases.

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