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SCHOLIUM. When the point A is at one of the extremities B, of the given line BC, any right line, drawn from that point to the circumference of the circle CEF, will be the one required.

PRO P. III. PROBLEM.

From the greater of two given right lines, to cut off a part equal to the less.

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Let AB and c be the two given right lines; it is required to cut off a part from AB, the greater, equal to c the lefs.

From the point A draw the right line AD equal to c (Prop. 2.); and from the centre A, at the distance AD, defcribe the circle DEF (Pof. 3.) cutting AB in E, and AE will be equal to c as was required.

For, fince A is the centre of the circle EDF, Ae will be equal to AD (Def. 13.)

But c is equal to AD, by conftruction; therefore AE will also be equal to c (Ax. 1.)

Whence, from AB, the greater of the two given lines, there has been taken a part equal to c the lefs, which was to be done.

SCHOLIUM. When the two given lines are fo fituated, that one of the extremities of c falls in the point A, the former part of the conftruction becomes unneceffary.

PROP.

PROP. IV. THEOREM.

If two fides and the included angle of one triangle, be equal to two fides and the included angle of another, each to each, the triangles will be equal in all refpects.

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Let ABC, DEF be two triangles, having CA equal to FD, CB to FE, and the angle c to the angle F; then will the two triangles be equal in all respects.

For conceive the triangle ABC to be applied to the triangle DEF, fo that the point c may coincide with the point F, and the fide CA with the fide FD.

Then, because CA coincides with FD, and the angle c is equal to the angle F (by Hyp.), the fide CB will alfo coincide with the fide FE.

And, fince CA is equal to FD, and CB to FE (by Hyp.), the point A will fall upon the point D, and the point B upon the point E.

But right lines, which have the same extremities, muft coincide, or otherwise their parts would not lie in the fame direction, which is abfurd (Def. 5.); therefore AB falls upon, and is equal to DE.

And, because the triangle ABC is coincident with the triangle DEF, the angle A will be equal to the angle D, the angle в to the angle E, and the two triangles will be equal in all respects (Ax. 9.) Q. E. D.

PROP. V. THEOREM.

The angles at the base of an ifofceles triangle are equal to each other.

D

C

Let ABC be an isofceles triangle, having the fide CA equal to the fide CB; then will the angle CAB be equal to the angle CBA.

For, in CA and cв produced, take any two equal parts CD, CE (Prop. 3), and join the points AE, BD:

Then, because the two fides CA, CE of the triangle CAE, are equal to the two fides CB, CD of the triangle CBD, and the angle c is commòn, the fide AE will also be equal to the fide BD, the angle CAE to the angle CBD, and the angle D to the angle E (Prop. 4.)

And fince the whole CD is equal to the whole CE (by Conft.), and the part CA to the part CB (by Hyp.), the remaining part AD will alfo be equal to the remaining part BE (Ax. 3.)

The two fides DA, DB, of the triangle DAB, being, therefore, equal to the two fides EB, EA of the triangle EBA, and the angle D to the angle E, the angle ABD will alfo be equal to the angle BAE (Prop. 4.)

And if from the equal angles CAE, CBD, there be taken the equal angles BAE, ABD, the remaining angle

CAB

CAB will be equal to the remaining angle CBA (Ax. 3.) Q. E. D.

COROLLARY. Every equilateral triangle is alfo equiangular.

PROP. VI. THEOREM.

If two angles of a triangle be equal to each other, the fides which are opposite to them will also be equal.

D

B

Let ABC be a triangle, having the angle CAB equal to the angle CBA; then will the fide CA be equal to the fide CB.

For if CA be not equal to CB, one of them must be greater than the other; let CA be the greater, and make AD equal to BC (Prop. 3.), and join BD.

Then, because the two fides AD, AB, of the triangle ADB, are equal to the two fides BC, BA, of the triangle ACE, and the angle DAB is equal to the angle CBA (by Hyp.), the triangle ADB will be equal to the triangle ACB (Prop. 4.), the lefs to the greater, which is abfurd.

The fide CA, therefore, cannot be greater than the fide CB; and, in the fame manner, it may be shewn that it cannot be lefs; confequently they are equal to each other. QE. D.

COROL. Every equiangular triangle is also equilateral.

PROP. VII. THEOREM.

If the three fides of one triangle be equal to the three fides of another, each to each, the angles which are oppofite to the equal fides will also be equal.

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Let ABC, DEF be two triangles, having the fide AB equal to the fide DE, AC to DF, and BC to EF; then will the angle ACB be equal to the angle DFE, BAC to EDF, and ABC to DEF.

For, let the triangle DFE be applied to the triangle ACB, fo that their longeft fides, DE, AB, may coincide, and the point F fall at G; and join CG.

Then, fince the fide AC is equal to the fide DF, or AG (by Hyp.), the angle ACG will be equal to the angle AGC (Prop. 5.)

And, because the fide BC is equal to the fide EF, or BG (by Hyp.), the angle BCG will be equal to the angle BGC (Prop. 5.)

But fince the angles ACG, BCG are equal to the angles AGC, BGC, each to each, the whole angle ACB will be equal to the whole angle AGB (Ax. 8.)

And, because AC is equal to AG, BC to BG, and the angle ACB to the angle AGB, the angle CAB will, also,

be

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