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But the angle ACE has been fhewn to be equal to the angle BAD; whence the angle ACE is also equal to the angle AEC; and the fide AE to the fide AC (I. 5.)

And, fince BEC is a triangle, and AD is drawn parallel to the fide EC, BD will be to DC as BA is to AE (VI. 3.); or, because AE is equal to AC, BD will be to DC as BA is to AC.

Again, let BD be to DC as BA is to AC; then will the angle BAC be bifected by the line AD.

For, let the fame construction be made as before:

Then fince BD is to DC as BA is to AC, (by Hyp.), and BD to DC as BA to AE (VI. 3.), therefore, also, BA will be to AC as BA is to AE.

And fince magnitudes which have the fame ratio to the fame magnitude are equal to each other (V. 10.), AC will be equal to AE, and the angle AEC to the angle ACE (I. 5.)

But the angle AEC is equal to the outward oppofite angle BAD (I. 28.); and the angle ACE is equal to the alternate angle CAD (I. 24.) ; whence the angle BAD will alfo be equal to the angle CAD.

Q. E. D.

PROP. V. THEOREM.

The fides about the equal angles of equiangular triangles are proportional; and if the fides about each of their angles be proportional, the triangles will be equiangular.

G

Let ABC, DEF be two equiangular triangles, of which BAC, EDF are correfponding angles; then will the fide AB be to the fide AC, as the fide DE is to the fide DF.

For make AG equal to DE, and AH to DF (I. 3.); and join GH:

Then, fince the two fides AG, AH of the triangle AHG, are equal to the two fides DE, DF of the triangle DFE, and the angle A to the angle D, the angle AGH will also be equal to the angle DEF (I. 4.)

But the angle DEF is equal to the angle ABC (byt Hyp.); confequently the angle AGH will also be equal to the angle ABC, and GH will be parallel to BC (I. 23.)

And, fince the line GH is parallel to the line BC, the fide AB will be to the fide AC, as the fide AG is to the fide AH (VI. 3.)

But AG is equal to DE, and AH to DF; whence the fide AB will be to the fide AC, as the fide DE is to the fide DF.

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Again, let AB be to AC, as DE is to DF; and AB to BC, as DE to EF; then will the triangle ABC be equiangular with the triangle DEF.

For, let the fame conftruction be made as before:

Then, fince AB is to AC as AG is to AH (by Hyp.), the line GH will be parallel to the line вс (VI. 3.), and the triangle AGH will be equiangular with the triangle ABC.

And fince, the fides about the equal angles of equiangular triangles are proportional (VI. 5.), the fide AB will be to the fide BC, as the fide AG is to the fide GH.

But the fide AB is to the fide BC, as the fide DE is to fide EF (by Hyp.); therefore, alfo, the fide AG will be to the fide GH, as the fide DE is to the fide EF (V. 11.).

And, fince the fide AG is equal to the fide DE (by Conft.), the fide GH will also be equal to the fide EF (V. 10.), and confequently the triangle DEF will be equiangular with the triangle AGH (I. 7.) or ABC, as was to be fhewn.

PRO P. VI. THEOREM.

If two triangles have one angle of the one equal to one angle of the other, and the fides about the equal angles proportional, the triangles will be equiangular.

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Let ABC, DEF be two triangles, having the angle A equal to the angle D, and the fide AB to the fide AC, as

the fide DE is to the DF; then will the triangle ABC be equiangular with the triangle DEF.

For, make AG equal to DE, and AH to DF; and join GH:

Then, fince the fides AG, AH are equal to the fides DE, DF, and the angle A to the angle D (by Hyp.), the fide GH will alfo be equal to the fide EF, and the triangle AGH to the triangle DFF (I. 4.)

And, fince AB is to AC as AG is to AH (by Hyp.), the line GH will be parallel to the line вC (VI. 3.); and confequently the angle AGH is equal to the angle ABC, and the angle AHG to the angle ACB (I. 25.)

The triangle ABC is, therefore, equiangular with the triangle AGH, and confequently it will alfo be equiangular with the triangle DEF. Q. E. D.

PRO P. VII. THEOREM.

In a right angled triangle, a perpendicular from the right angle, is a mean proportional between the fegments of the hypotenufe; and each of the fides is a mean proportional between the adjacent segment and the hypotenuse.

D

Let ABC be a right angled triangle, and CD a perpendicular from the right angle to the hypotenuse; then will

CD be a mean proportional between AD and DB; Aca mean proportional between AB and AD; and BC between AB and BD.

For, fince the angle BDC is equal to the angle ACB (I. 8.), and the angle в is common, the triangles DBC, ABC will be equiangular (I. 28. Cor.)

And, in the fame manner, it may be fhewn, that the triangles ADC, ABC are equiangular; whence the triangle ADC is alfo equiangular with the triangle DBC.

But the fides of equiangular triangles are proportional (VI. 5.); therefore the fide AD is to the fide CD, as the fide CD is to the fide DB.

In like manner, fince the triangles ADC, DBC are each of them equiangular 'with the triangle ABC, AB will be to AC as AC is to AD; and AB to BC as BC is to BD (VI. 5.)

Q. E. D.

COROLL. If ACB be a femicircle, and CD a perpendicular let fall from any point c; then will ADXDB-DC2, ABXAD AC, and ABXBDBC2.

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