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If two sides of one triangle be respectively equal to those of another, but stand on a greater base; the angle contained by the former will be likewise greater than what is contained by the latter.

Let the triangles ABC and DEF have the sides AB and BC equal to DE and EF, but the base AC greater than DF; the vertical angle ABC is greater than DEF.

For if ABC be not greater than the angle DEF, it must

either be equal or less. But it - ic

Jo cannot be equal to DEF, for the sides AB, BC being then equal to DE, EF, and contain

ing equal angles, the base AC A. C in F would (I. 3.) be equal to DF, which is contrary to the hypothesis. Still more absurd it would be to suppose the angle ABC less than DEF, since the triangles BAC and EDF, having their sides AB, BC equal to DE, EF, but the contained angle ABC less than DEF, or DEF greater than ABC, the base DF would, from the preceding pro

position, be greater than AC, or AC would be less than IDF. -


Two triangles are equal, which have two angles and a corresponding side in the one respectively. equal to those in the other. *

Let the triangles ABC and DEF have the angle BAC equal to EDF, the angle BCA to EFD, and a side of the one equal to a side of the other, whether it be interjacent or opposite to those equal angles; the triangles will be equal.

First, let the equal sides be AC and DF, which are interjacent to the equal angles in both triangles.—Apply the triangle ABC to DEF; the point A being laid on D, and the straight line AC on DF, the other extremities C and F must coincide, since those lines

are equal. And because the angle B BAC is equal to EDF, and the IE side AC is applied to DF, the other side AB must lie along DE; A. C ID 15

and for the same reason, the angles BCA and EFD being equal, the side CB must lie along FE. Wherefore the point B, which is common to both the lines AB and CB, will be found likewise in both DE and FE; that is, it must fall upon the corresponding vertex E. The two triangles ABC and DEF, thus mutually adapting, are hence entirely equalNext, let the equal sides be AB and DE, which are opposite to the equal angles BCA and EFD. The triangle ABC being laid on DEF, the sides AB and AC of the angle BAC will apply to DE and DF, Io.

the sides of the equal angle EDF; 4. T} and since AB is equal to DE, the points B and E must coincide; but

by hypothesis, the angles BCA and A. C D I' EFD being equal, BC must adapt itself to EF, for otherwise one of those angles becoming the exterior angle of a secondary triangle, would (I. 8.) be greater than the other.

Whence the triangles ABC, DEF are entirely coincident, and have those sides equal which subtend equal angles. Schol. By the application of the first case, where the sides lying between the equal angles are equal, the distance of an inaccessible object can be measured in all cases.


Two triangles are equal if two sides and a corresponding opposite angle be equal in both, and the other opposite angles have the same character.

In the triangles ABC and DEF, let the side AB be equal to DE, BC to EF, and the angles BAC, EDF, opposite to BC, EF, be also equal; the triangles themselves are equal, if the other angles BCA and EFD opposite to AB and DE be of the same character, or at once right, or acute, or obtuse.

For, the triangle ABC being applied to DEF, the an

gle BAC will adapt itself to EDF, since they are equal; and the point B must coincide with E, because the side AB is equal to DE. But the other equal sides BC and EF, now stretching from the same point

E towards DF, must likewise coin- To TE cide; for if the angle at C or F be /\ Z right, there can exist no more than

- 4- CID F one perpendicular EF (I. 17. cor.) and, in like manner, if this angle at F be either obtuse or acute, the line EF, which forms it, can, for the same reason, have only one corresponding position.—Whence, in

each of these three cases, the triangle ABC admits of a perfect adaptation with DEF.

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If a straight line fall upon two parallel straight lines, it will make the alternate angles equal, the exterior angle equal to the interior opposite one, and the two interior angles on the same side together equal to two right angles.

Let the straight line EFG fall upon the parallels AB and CD ; the alternate angles AGF and DFG are equal, the exterior angle EFC is equal to the interior angle EGA, and the interior angles CFG and AGF, or FGB and GFD, are together equal to two right angles.

For conceive a straight line, produced both ways from F, to turn about that point in the same plane; it will first cut the extended line AB above G and towards A, and will in its progress afterwards meet this line on the other side below G and towards B. In the position IFH, the angle EFH is the exterior angle of the triangle FHG, and therefore greater than FGH or EGA (I. 8.) But in the last position LFK, the exterior angle EFL is equal to its vertical angle GFK in the triangle FKG, and to which the angle FGA is exterior ; consequently (I. 8.) FGA is greater than EFL, or the angle EFL is less than FGA or EGA. When the incident line EFG, therefore, meets AB above the point G, it makes an angle EFH greater than EGA; K. and when it meets AB below that B point, it makes an angle EFL, which is less than the same angle. But in passing through

4/ C/ L.


all the degrees from greater to less, a varying magnitude must evidently encounter the single intermediate limit of equality. Wherefore, there is a certain position CD, in which the line revolving about the point F makes the exterior angle EFC equal to the interior EGA, and at the same instant of time meets AB neither towards the one part nor the other, or is parallel to it.

And now, since EFC is proved to be equal to EGA,

and is also equal to the vertical angle GFD; the alternate angles FGA and GFD are equal. Again, because GFD and FGA are equal, add the angle FGB to each, and the two angles GFD and FGB are equal to FGA and FGB; but the angles FGA and FGB, on the same side of AB, are equal to two right angles, and consequently the interior angles GFD and FGB are likewise equal to two right angles. Cor. Since the position CD is individual, or that only one straight line can be drawn through the point F parallel to AB, it follows that the converse of the proposition is likewise true, and that those three properties of parallel lines are criteria for distinguishing parallels.


Through a given point, to draw a straight line parallel to a given straight line. -

To draw, through the point C, a straight line parallel to AB. In AB take any point D, join CD,

and at the point C make (I. 4.) an an- TST. gle DCE equal to CDA ; CE is paral- x lel to AB.


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