« AnteriorContinuar »
And, since ratios which are the fame to the fame ratio, are the same to each other (V. 11.), DB will be to DA as Ec is to EA; or, inversely, AD to DB as Ae to ec.
Again, let the sides AB, AC be cut proportionally, in the points D and E; then will the line De be parallel
For, 'the same construction being made as before, the triangle des will be to the triangle Dea, as DB is to DA (VI. 1.); and the triangle Edc to the triangle dea as EC EA (VI. 1.)
And, since DB is to da as Ec is to EA (by Conft.), the triangle der will be to the triangle DEA as the triangle EDC is to the triangle dea (V. 11.)
But magnitudes which have the same ratio to the fame magnitude are equal to each other (V. 10.); whence the triangle DEB is equal to the triangle EDC.
And since these triangles are equal to each other, and are upon the same base de, they will have equal altitudes (VI. 2. Cor.), or stand between the fame parallels; whence DE is parallel to BC.
Q. E. D. COROLL. In the same manner it may be thewn, that the sides of the triangle are proportional to any two of the parts into which they are divided; and that the like parts of each are alfo proportional.
If the vertical angle of a triangle be bisected, the segments of the base will have the same ratio with the other two sides : and if the segments have the same ratio with the other two fides the angle will be bifected.
Let the angle BAC of the triangle ABC be bisected by the right line AD; then will be be to dc as BA is to AC.
For through the point c draw ce parallel to DA (I. 27.), and let Ba be produced to meet ce in E:
Then, because the right line Ac cuts the two parallel right lines AD, ec, the angle Ace will be equal to the alternate angle CAD (I. 24.)
But the angle cad is equal to the angle BAD, by the propofition; therefore the angle Bad is also equal to the angle ACE.
And, in like manner, because the right line Be cuts the two parallel right lines AD, EC, the outward angle BAD will be equal to the inward opposite angle AEC (I. 25.) M 2
But the angle ACE has been shewn to be equal to the angle BAD ; whence the angle Açe is also equal to the angle AEC ; and the side Ae to the side AC (I. 5.)
And, since bec is a triangle, and an is drawn parallel to the side ec, BD will be to dc as ba is to AE (VI. 3.) ; or, because he is equal to AC, BD will be to do 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 same construction be made as before :
Then fince bp 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 since 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 opposite angle BAD (I. 28.); and the angle ace is equal to the alternate angle CAD (I. 24.) ; whence the angle BAD will also be equal to the angle CAD.
Q. E, D.
PRO P. V. THEOREM.
The sides about the equal angles of equiangular triangles are proportional; and if the sides about each of their angles be proportional, the triangles will be equiangular.
Let ABC, DEF be two equiangular triangles, of which BAC, EDF are corresponding angles; then will the side AB be to the fide ac, as the fide de is to the
For make AG equal to De, and Ah to DF:(I. 3.); and join th:
Then, since the two fides AG, AH of the triangle AHG, are equal to the two sides DE, DF of the triangle Die, and the angle A to the angle D, the angle AGH will also be equal to the angle det (I. 4.)
But the angle DEF is equal to the angle ABC (by Hyp.); consequently the angłe AGH will also be equal to the angle ABC, and GH will be parallel to BC (I. 23.)
And, since the line GH is parallel to the line BC, the fide AB will be to the side ac, as the side AG is to the side Ah (VI. 3.)
But AG is equal to DE, and Ah to DF; whence the fide AB will be to the side AC, as the fide DE is to the side DF.
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 same construction 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 bc (VI. 3.), and the triangle AGH will be equiangular with the triangle ABC.
And since the sides about the equal angles of equi. angular triangles are proportional (VI. 5.), the side AB will be to the side BC, as the side AG is to the side GH.
But the side AB is to the side BC, as the fide de is to fide EF (by Hyp.); therefore, also, the side AG will be to the side GH, as the fide de is to the side er (V. 11.).
And, since the side ag is equal to the side de (by Conft.), the fide GH will also be equal to the side er (V. 10.), and consequently the triangle DEF will be equiangular with the triangle AGH (1. 7.) or ABC, as was to be shewn.
PROP. VI. THEOREM.
If two triangles have one angle of the one equal to one angle of the other, and the sides about the equal angles proportional, the triangles will be equiangular.
Let ABC, der be two triangles, having the angle A equal to the angle D, and the side AB to the side Ac, as