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the fide de is to the DF; then will the triangle ABC be equiangular with the triangle der.

For, make AG equal to de, and ah to DF; and join GH:

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

And, since AB is to AC as AG is to Ah (by Hyp.), the line GH will be parallel to the line bc (VI. 3.); and con{equently the angle AGH is equal to the angle ABC, and the angle ang to the angle ACB (I. 25.)

The triangle abc is, therefore, equiangular with the triangle Agh, and consequently it will also be equiangular with the triangle DEF.

Q. E, D,

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PRO P. VII. THEOREM.

In a right angled triangle, a perpendicu. lar from the right angle, is a mean proportional between the segments of the hypotenuse ; and each of the sides is a mean proportional between the adjacent segment and the hypotenuse.

Let ABC be a right angled triangle, and co a perpendicular from the right angle to the hypotenuse ; then will cd be a mean proportional between AD and DB ; AC a mean proportional between AB and AD; and BC between AB and BD.

For, since the angle bbc is equal to the angle ACB (I. 8.), and the angle B is common, the triangles DBC, ABC will be equiangular (I. 28. Cor.)

And, in the same manner, it may be shewn, that the triangles ADC, ABC are equiangular ; whence the triangle adc is also equiangular with the triangle DBC.

But the sides of equiangular triangles are proportional (VI. 5.); therefore the fide ad is to the side CD, as the fide cd is to the side DB.

In like manner, since 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 semicircle, and cd a perpendicular let fall from any point c; then will ADXDB=pc", ABXAD=AC, and ABXBD=BC?,

PROP, VIII. PROBLEM,

To find a third proportional to two given right lines.

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Let A, B be two given right. lines ; it is required to find a third proportional to them.

Draw the two indefinite right lines CF, cg, making any angle c, and take cd equal toʻA, and ce, df each equal B (I. 3.)

Also join the points D, E, and make fg parallel to DE (I. 27.) ; and EG will be the third proportional required.

For, fince CFG is a triangle, and dE is parallel to FG (by Conft.) cd will be to DF as ce is to EG (VI. 3.)

But DF is equal to ce, by construction; therefore cd is to ce as ce is to eG.

And, since cd is equal to A, and ce to B (by Conft.), A will be to B as B

Q. E. D.

to EG.

SCHOLIUM. A third proportional to two given right ļines, may also be found by means of the last propos sition.

For let AD, DC be two given right lines, one of which pc is perpendicular to the other.

Then Then if co be drawn at right angles to Ac, and AD be produced till it meets the former in , DB will be a third proportional to AD and Dc as was required.

Again, let BD, DC be the two given right lines, placed at right angles to each other, as before :

Then, if ca be drawn at right angles to BC, and BD be produced till it meets the former in A, DA will be a fourth proportional to BD and oc.

PRO P. IX. PROBLEM.

To find a fourth proportional to three given right lines.

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Let A, B, C be three given right lines ; it is required to find a fourth proportional to them.

Draw the two indefinite right lines DG, DH, making any angle d; and take de equal to A, DF to B, and eG to c (I. 3.)

Join the points E, F, and make Gh parallel to EF (1.27.); and Fh will be the fourth proportional required.

For, since DGH is a triangle, and EF is parallel to GH (by Conft.) DE will be to DF as EG is to FH (VI. 3.)

But de is equal to A, DF to B, and Eg to c (by Confi.); therefore a will be to B as c is to Fh (V. 9.)

Q. E. D.

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To find a mean proportional between two given right lines.

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Let A, B be two given right lines; it is required to find a mean proportional between them.

Draw the indefinite right line ca, in which'take ca. equal to A, and ED equal to B (I. 3.)

Upon cd describe the semi-circle CFD, and from the point Ę erect the perpendicular EF (1. 11.); and it will be the mean proportional required.

For join the points CF, FD :

Then, because DFC is a semi-circle, the angle crd is a right angle (III. 16.), and consequently the triangle cdF is rectangular.

And since a perpendicular from the right angle is a mean proportional between the segments of the hypotenuse (VI. 7.), EF will be a mean proportional to ce,

ED.

But ce is equal to A, and Ed to B (by Conft.); therefore EF will be a mean proportional to A and B, as was to be fhewn.

SCHOLIUM. If CD, De be the two given lines, DF will be a mean proportional to them..

And if cD, ce be the given lines, ce will be a mean proportional to them (Vl. 7.)

PROP.

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