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Book VI.

COR. The triangle ADE is fimilar to the triangle ABC. For the two triangles BAD, CAE having the angles at D and E right angles, and the angle at A common, are equiangular, and therefore BA: AD:: CA: AE, and alternately BACA::AD: AE; therefore the two triangles BAC, DAE, have the angle at A common, and the fides h 6. 6. about that angle proportionals, therefore they are equiangular h and fimilar.

Hence the rectangles BA.AE, CA.AD are equal.

[blocks in formation]

F from any angle of a triangle a perpendicular be drawn to the oppofite fide, or bafe; the rectangle contained by the fum and difference of the other two fides, is equal to the rectangle contained by the fum and difference of the segments, into which the base is divided by the perpendicular.

Let ABC be a triangle, AD a perpendicular drawn from the angle A on the base BC, fo that BD, DC are the segments of the bafe; AC+AB. AC—AB⇒CD+DB. CD-DB.

[blocks in formation]

From A as a centre with the radius AC, the greater of the two fides, defcribe the circle CFG; produce AB to meet the circumference in E and F, and CB to meet it in G. Then because AF AC, BF AB+AC, the fum of the fides; and Gince AE AC, BE AC-AB the difference of the fides. Also,

Alfo, because AD drawn from the centre cuts GC at right Book VI.
angles, it bisects it; therefore, when the perpendicular falls
within the triangle, BG DG-DB=DC-DB the dif-
ference of the fegments of the base, and BC=BD+DC= the
fum of the fegments. But when AD falls without the tri-
angle, BG=DG+DB=CD+DB the fum of the fegments
of the bafe, and BC CD-DB the difference of the feg-
ments of the base. Now, in both cafes, because B is the in-
terfection of the two lines FE, GC, drawn in the circle,
FB.BE CB.BG; that is, as has been shown,

AC+AB. AC—AB=CD+DB. CD-DB. Therefore, &c.
Q. E. D.

SUPPLEMENT

SUPPLEMENT

TO THE

ELEMENTS

OF

GEOMETRY.

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