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Let ABC be a circle, of which AC is a diameter ; let DE Book VI. be perpendicular to the diameter AC, and let AB meet DE in F; the rectangle BA.AF is equal to the rectangle CA.AD.

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Join BC; then the angle ABC is a right anglea. Now the a 31. 3. angle ADF is a right angleb; and the angle BAC is either b Hyp. the same with DAF, or vertical to it; therefore the triangles ABC, ADF are equiangular, therefore BA : AC :: AD : AFc; therefore the rectangle BA.AF is equal to the rect- c 4.6. angle AC.ADA. Therefore, if from the extremity, &c. d 16.6. Q. E. D.

PROP. H. THEOR.

THE perpendiculars drawn from the three angles of any triangle to the opposite sides intersect one another in the same point.

Let ABC be a triangle, and BD, CE two perpendiculars intersecting each other in F; let AF be joined, and produced, if necessary, to meet BC in G; AG is perpendicular to BC.

Book VI.

a 5. 4.

31. 3.

Join DE, and about the triangle AEF, describe a circle
AEF; then, because AEF is a right angle, the circle de-
scribed about the triangle
AEF will have AF for its
diameterb. In the same
manner the circle de-
scribed about the triangle
ADF has AF for its dia-
meter. Therefore the

D
points A, E, F, D are in
the circumference of the E
same circle. Because the
angle EFB is equal to the
angle DFC, and the an-
gle BEF to the angle

B
CDF, the triangles BEF,
CDF are equiangular; therefore BF: EF :: CF: FDd, or,
alternatelye, BF : FC : : EF : FD. Therefore since the
sides about the equal angles BFC, EFD are proportional,
the triangles BFC, EFĎ are equiangularf, and the angle
FCB is equal to the angle EDF. But EDF is equal to
EAFs. Therefore the angle EAF is equal to the angle
FCG. Now the angles AFE, CFG are equale; therefore
the remaining angles AEF, FGC are equall. But AEF is
a right angle, therefore FGC is a right angle ; therefore AG
is perpendicular to BC. Q. E. D.

c 15. 1.

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d 4. 6. e 16. 5.

f 6. 6.

8 21.3.

h 32. 1.

PROP. K. THEOR.

IF perpendiculars be drawn from any two angles of a triangle to the opposite sides, the rectangles contained by the two sides on which the perpendiculars fall and the segments intercepted between the perpendiculars and the angle opposite to the third side will be equal. See the last figure.

Let ABC be a triangle, and BD, CE two perpendiculars from the angles at B, C to the opposite sides AC, AB; the rectangles AB.AE and AC.AD are equal.

For the triangles BAD, CAE, having the angles at D and Book VI. E right angles, and the angle at A common, are equiangular ; therefore BA : AD:: CA: AE, and alternatelyb, BA : CA a 4.6. :: AD: AE; therefore the rectangles BA.AE, CA.AD are b 16. 5. equal.

Cor. Hence the triangles ABC, ADE, having their sides about the common angle A proportional, are equiangular and similar.

C 6.6.

PROP. L. THEOR.

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IF from any angle of a triangle a perpendicular be drawn to the opposite side or base, the rectangle contained by the sum and difference of the other two sides is equal to the rectangle contained by the sum and difference of the segments into which the base is divided by the perpendicular.

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Let ABC be a triangle, and AD a perpendicular drawn
from the angle A on the base BC, so that BD, DC are the seg-
ments of the base ;
AC+AB.AC-AB=CD+DB.CD_DB.

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15 the

From A as a centre, with the radius AC, the greater of the two sides, describe the circle CFG; produce AB to meet the circumference in E, F, and CB to meet it in G. Because

a 3. 3.

Book VI. AF=AC, BF=AB+AC=the sum of the sides ; and since

AE=AC, BE=AC-AB=the difference of the sides. AD bisects GCa, therefore, when the perpendicular AD falls within the triangle, BG=DG-DB=DC-DB=the difference of the segments of the base, and BC=BD+DC=the sum of the segments; but when AD falls without the triangle, BG=DG+DB=CD+DB=the sum of the segments

of the base, and BC=CD-DB==the difference of the segb 35. 3. ments. Now, in both cases, FB.BE=CB.BG”, that is, as

has been shown,
AC+AB. ACAB=CD+DB.CD-DB.
Therefore, if from any angle, &c. Q. E. D.

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