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DEM. Draw CG, DH, parallel respectively to AB, FE. By ART. 24, the parallelograms ABGC, DHEF, are equal. Hence, by ART. 22, the triangles ABC, DEF, are also equal*. This, &c.

ART. 27. Equal triangles upon the same base, and upon the same side of it, are between the same parallels.

Let ABC, ADC, be two equal B triangles on the same base AC, and on the same side of it. Then these triangles are between the same parallels.

DEM. Draw the right line BD from the vertex of the one triangle to the vertex of the other: BD is parallel to AC. For, suppose that BD is not parallel to AC, but that any other line, as BE, meeting a side AD of the triangle ADC in the point E is parallel to AC. Then, drawing from this point a right line to the vertex of the opposite angle c, the triangle AEC would be equal to the triangle ABC, by ART. 25; and consequently to the triangle ADC which is given equal to ABC. But this is impossible; and therefore the supposition is false: BE is not parallel to AC. Hence, as it may be proved in the same manner that no other line except BD is parallel to AC, BD is parallel to AC. [See NOTE M.]

This, &c.

DEF. XII. Taking any side of a triangle as base, the perpendicular from the vertex to the base (produced if necessary), is called the Altitude of the triangle. Also, taking any side of a parallelogram as base, the perpendicular from any point in the opposite side, on the base (produced if necessary), is called the Altitude of the parallelogram,

B

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Thus, if ABC, DEFG, be a triangle and parallelogram, on their respective bases AC, DG; and if BH, IK, be perpendiculars to AC, DG; then BH, IK are respectively the altitudes of ABC, defg.

A

H C

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* By Ax. 7.

C

This perpendicular, it is evident, may be taken any where in the Aparellel EF, because it is always the same length. For if AB, cd,

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be parallel, any two perpendiculars, such as IK, LM, are equal. For by ART. 17, IK and LM are parallel, and therefore, by ART. 20, they are equal, as IKML is a parallelogram.

ART. 28. Parallelograms which have equal bases and equal altitudes are equal.

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DEM DRAW AL, EM parallel respectively to DI, HK; and produce IB, KF, to meet them in L and M. The figures ALID, EMKH, are parallelograms by DEF. XI. Now, if the diagonals IA, KE be drawn, the triangles IDA, KHE, are equal, by ART. I; because they have two sides, Id, da, granted respectively equal to two, KH, HE, and the contained angles IDA, KHE also equal, by ART. 8. Consequently ALID, EMKH, the doubles of these triangles, are also equal*. Hence, inasmuch as, by ART. 23, ALID equals ABCD, and EMKH equals EFGH, the parallelograms ABCD, EFGH, are likewise equal. This, &c.

ART. 29. Triangles which have equal bases and equal altitudes are equal.

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DEM. Draw AI, BI, parallel to CB, CA respectively; and

Ax. 8. The doubles of equal things are equal.

DK, EK parallel to FE, FD respectively. Then, by ART. preceding, the parallelograms AIBC, DKEF are equal, having equal bases AC, DF, and equal altitudes BG, EH. Hence, by ART. 22, the triangles ABC, DEF, their halves, are also equal. This, &c.

ART 30. Equal triangles on equal bases have equal altitudes.

B

A G

E

K

M

D

H F

Let ABC, DEF, be two equal triangles, upon equal bases, AC and DF. Then their altitudes, BG and EH, are also equal. DEM. For suppose EH to be greater than BG, and that HL is equal to BG. Draw BI, CI, parallel respectively to AC, AB; and EK, FK, parallel respectively to DF, DE; also, through the point L draw MN parallel to DF. Then, by ART. 22, the parallelograms ABIC, DEKF are equal, because their halves, the triangles, are given equal t. But, by ART. 28, the parallelogram ABIC would, on the above supposition, be also equal to the parallelogram DMNF; and therefore DMNF would be equal to DEKF,-a part to the whole, which is impossible. Hence this supposition is false; EH is not greater than BG. In the same way, it may be proved that BG is not greater than EH. Hence BG and EH are equal. This, &c.

ART. 31. If a parallelogram and a triangle be upon the same base, and between the same parallels, the paral· lelogram is double of the triangle.

Let ACED be the parallelogram on the same base AC as the triangle ABC, and between the same parallels AC, BE. Then ACED is double of ABC.

DEM.

Draw the diagonal DC.

B

D

E

A

By ART. 25, the triangles ABC, ADC are equal; and by

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ART. 22, ACED is double of ADC. Hence, ACED is double of ABC *. This, &c.

DEF. XIII. A square is a parallelogram whose two adjacent sides are equal, and any of whose angles is a right angle.

Thus, if in the parallelogram ACBD, any pair of adjacent sides AC, AD be equal, and any one angle ACB be a right angle, ACBD is called a square. [See NOTE N.]

B

D

ART. 32. All the sides of a square are equal, and all its angles right angles.

By DEF. XIII. and ART. 20, all its sides are equal. By ART. 13, the angle CBD is a right one, since ACB is a right one; and therefore, by ART. 21, all the angles are right. This, &c.

ART. 33. Squares described upon equal right lines are equal.

Let ABCD, EFGH be two squares described upon the equal right lines AD, EH. Then these squares are equal.

This is evident. [See NOTE O.]

B C

F

G

A

D E

H

PROB. X. On a given right line to describe a square. Let AB be the given line. It is required to describe a square on it.

CONS. At either extremity of the given line, as A, raise the perpendicular Ac, by PROB. VII., and take ac equal to AB, by PROB. III. Through the points в and c draw the right lines BD, CD, parallel respectively to ca and AB, by PROB. IX.

C

D

A

B

and meeting in D. Then the figure ACDB is a square.

DEM. BY DEF. XI., ACDB is a parallelogram; and as Ac equals AB, and the angle at A is a right angle, ACDB is a square by DEF. XIII. This, &c.

* By Ax. 8.

To see how a square is described upon a given right line, read Problems to number X.

ART. 34. If two squares be equal, their sides are equal. Let the squares (in fig. above) ABCD, EFGH be equal. Then also their sides AD, EH are equal.

This is evident. [See NOTE P.]

LESSON V.

ART. 35. If any side of a triangle be produced, the external angle is equal to the two farther internal angles taken together.

Let ABC be any triangle, of which the side AC is produced. Then the external angle BCD is equal to the two farther internal angles ABC and BAC.

DEM. Draw CE parallel to AB. By A

B

E

C D

ART. 12, the angle ABC is equal to вCE; and by Art. 14, the angle BAC is equal to ECD. Hence the two angles ABC, BAC together are equal to the two angles BCE, ECD together, that is, to the whole angle BCD. This, &c.

ART. 36. The external angle of any triangle is greater than either of the two farther internal angles.

This is immediately evident from the last Article. ART. 37. The three internal angles of any triangle taken together, are equal to two right angles.

In the triangle ABC, the angles ABC, BCA, CAB, taken together, are equal to two right angles.

DEM. BY ART. 9, the angles BCA, BCD together are equal to two right angles; but by ART. 35, the angle BCD is equal to the two angles ABC, BAC together. Hence the angles BCA, ABC, BAC together are equal to A two right angles. This, &c.

B

C

D

ART. 38. Any two angles of a triangle are together less than two right angles; and if any angle of a triangle be obtuse or right, the other two are acute; also, if two angles of a triangle be equal, they are both acute.

These are immediately evident from the last Article.

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