« ForrigeFortsett »
Book III. is bifected in E, and produced to the point D, the rectangle AD, WDC, together with the square of EC, is equal to the square b 6.2. of ED, and CE is equal to EB: There
fore the rectangle AD, DC, together
square of ED: But the square of ED C 47. 1. is equal < to the squares of EB, BD, be
cause EBD is a right angle: Therefore
But if DCA does not pass through d 1. 3. the centre of the circle ABC, ked the centre E, and draw EF
perpendicular e to AC, and join EB, EC, ED: And because EF, which passes through the centre, cuts AC, which does not pass through the centre, at right angles,
D f 3. 3.
it shall bisect * it; therefore AF is equal
with the square of EB, is equal to the square of ED But C 47. 1. the squares of EB, BD are equal to the square of ED, be
cause EBD is a right angle; therefore the rectangle AD, DC, together with the square of EB, is equal to the squares of EB, BD: Take away the common square of EB; therefore the remaining rectangle AD, DC is equal to the square of DB. Wherefore, if from any point, &c. Q. E. D.
Cor. If from any point without a circle, there be drawn Book III. two straight lines cutting it, as AB, AC, the rectangles contained by the whole lines and the parts of them without the . circle, are equal to one another, viz.
I the rectangle BA, AF to the rectangle CA, AE: For each of them is equal to
D the square of the straight line AD which touches the circle.
PROP. XXXVII. THEOR.
two straight lines, one of which cuts the circle, and the other meets it; if the rectangle contained by the whole line which cuts the circle, and the
part of it without the circle be equal to the square of the line which meets it; the line which meets shall touch the circle.
Let any point D be taken without the circle ABC, and from it let two straight lines DCA and DB be drawn, of which DCA cuts the circle, and DB meets it; if the rectangle AD, DC be equal to the square of DB; DB touches the circle. Draw * the straight line DE touching the circle ABC, find
a 17.30 its centre F, and join FE, FB, FD; then FED is a right angle: And because DE touches the circle ABC, and DCA cuts it, the rectangle AD, DC is equal to the square of DE : But c 36. 3. the rectangle AD, DC is, by hypothesis, equal to the square of DB: Therefore the square of DE is equal to the square of DB; and the straight line DE equal to the straight line DB : And FÊ is equal to FB, wherefore DE, EF are equal to DB, BF; and the base FD is common to the B two triangles DEF, DBF; therefore the angle DEF is equal to the angle
d 8. I. DBF; but DEF is a right angle, there
F fore also DBF is a right angle: And FB, if produced, is a diameter, and the straight line which is drawn at right
Book III. angles to a diameter, from the extremity of it, touches the
circle: Therefore DB touches the circle ABC. Wherefore, if. € 16. 3. from a point, &c. Q. E. D.
Cor. Hence straight lines DB, DE, drawn to touch a circle ABC from the same point D without it, are equal to one, another.
PROP. A. THEOR.
together equal to two right angles, and the cir: cumference of a circle pass through three of these angles, it shall also pass through the fourth.
Let ABCD be a quadrilateral figure, of which the opposite angles are equal to two right angles, and let the circumference of a circle pass through the three points A, B, C; it shall alfo pass through the point D.
Join AČ, BD: and if the circumference ABC do not pass through D, let it, if possible, meet BD in the point E; and join AE, EC; therefore the
B. a 21. 1. angle AEC is unequal a to the angle
AĎG: but because ABCE is a quadri
lateral figure in a circle, the opposite b 22• 3. angles ABC, AEC are equal to two
right angles : and the two ABC, ADC
Cor. 1. And if two triangles ACD, ACE, on the same base
In the circumference take any point
AB, BC; therefore ABC, ADC are b 22. 3. equal to two right angles : but ADC
is equal to AEC; therefore ABC,
Cor. 2. If ADC be a right angle; the circumference of the Book III. circle, of which AC is the diameter, shall pass through the point D. Take any point B in the circumference on the other fide of AC, and join AB, BC; therefore the angle ABC in the semicircle is a right angle; and ABC, ADC are therefore equal c 31. 3. to two right angles. Wherefore the circumference ABC passes through the point D.
PROP. B. THEOR.
be equal to the rectangle contained by the segments of the other; the circumference which passes through three of their extremities, shall also pass through the fourth.
Let the two straight lines AB, CD cut one another in E; and let the rectangle AE, EB be equal to the rectangle CE, ED; also let the circumference of a circle pass through the three points A, C, B; it shall also pass through the point D.
If not, let it, if possible, cut CD in some other point F: and because the straight lines AB, CF in the circle cut one another in E, the rectangle AE, EB is equal a to the rectangle CE, C
D a 35. 3.
F EF: but the rectangle AE, EB is equal to the rectangle CE, ED; therefore the rectangle CE, EF is equal to the rectangle CE, ED; and EF is therefore equal to ED; the lefs to the greater, which is impossible. Therefore the circumference ACB must pass through the point D. Wherefore, &c. Q. E. D.
Cor. Likewise, if the straight line CD meet a circle in C, and meet a straight line AB, which cuts the circle in the point Ewithout it; and the rectangle CE, ED be equal to the rectangle AE, EB: it may be proved in the same manner, that the point D is in the circumference ; by using the Cor. to the 36th, instead of the 35th.
PROP. C. THEOR.
F a straight line be drawn from the extremity of
the diameter of a circle, and meet a perpendicular to that diameter; the rectangle contained by the segment of it within the circle, and the fegment between the extremity of the diameter and the perpendicular, is equal to the rectangle contained by the diameter, and the segment of it be. tween the same extremity and the perpendicular.
Let ABC be a circle, of which AC is a diameter, and let DE be perpendicular to AC, and from the extremity A, let AB be drawn, meeting the circumference again in B, and the perpendicular in E; the rectangle BA, AE is equal to the rectangle CA, AD.
Join BC, CE; and because ABC is an angle in a semicircle, a 31.3. it is a right angle &; therefore the circle of which CE is a dia. b 2. Cor. meter, shall pass through Bb: For the same reason, it shall
A. 3. pass through the point D: let this be the circle CBED: and
because the straight lines BE, CD, either cut one another in the
point A within the circle, or are drawn from the point A withC 35. 3. or out it; the rectangle BA, AE is equal to the rectangle CA, Cor. 36. 3. AD. Wherefore, &c. Q. E. D.
PROP. D. THEOR.
F from two angles of a triangle, perpendiculars
be drawn to the opposite sides, the straight line drawn from the third angle to the point where they meet, shall be perpendicular to the third fide of the triangle.