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tained by BE and EF, because EF is equal to ED ; therefore BD is equal to the square of EH; and BD is also equal to the rectilineal figure A ; therefore the rectilineal figure A is equal to the square of EH : Wherefore a square has been made equal to the given rectilineal figure A, viz. the square described upon EH. Which was to be done.
PROP. A. THEOR. If one side of a triangle be bisected, the sum of the squares of the other
two sides is double of the square of half the side bisected, and of the square of the line drawn from the point of bisection to the opposite angle of the triangle.
Let ABC be a triangle, of which the side BC is bisected in D, and DA drawn to the opposite angle ; the squares of BA and AC are together double of the squares of BD and DA.
From A draw AE perpendicular to BC, and because BEA is a right angle, AB2= (47. 1.) BE? + AE? and AC2=CE2 + ÀE?; wherefore AB2+ ACP=BE2 +CE2 +2AE2. But because the line BC is cut equally in D, and unequally in E, BE2 + ČE2 (9. 2.) 2BD2+2DE2 ; therefore AB +AC: == 2BD2 +2DE2 +2AE2.
Now DE+AE = (47. 1.) ADS, and 2DE2 + 2AE? =2AD? ; wherefore AB? + AC2 = 2BD2 + 2AD?..
D E Therefore, &c. Q. E. D.
PROP. B. THEOR.
The sum of the squares of the diameters of any parallelogram is equal to
the sum of the squares of the sides of the parallelogram. Let ABCD be a parallelogram, of which the diameters are AC and BD; the sum of the squares of AC and BD is equal to the sum of the squares of AB, BC, CD, DA.
Let AC and BD intersect one another in E: and because the vertical angles AED, CEB are equal (15. 1.), and also the alternate angles EAD, ECB (29. 1.), the triangles ADE, CEB have two angles in the one equal to two angles in the other, each to each : but the sides AD and BC, which are oppo
D site to equal angles in these triangles, are also equal (34. 1.) ; therefore the other sides which are opposite to the equal angles are also equal (26. 1.), viz. AE to EC, and ÈD to EB.
Since, therefore, BD is bisected in E, ABP +AD-= (A, 2.) 2BE2 +2AE2 ; and for the saine reason, CD2 +BC2=2BE2 +2EC2=2BE2 +2AEo, because EC=AE. Therefore AB2 +AD3 +DC? +BC2 = 4BE2 +4AE?. But 4BE2=BD2, and 4AE2=AC2 (2. Cor. 8. 2.) because BD and AC are both bisected in £; therefore ABS + AD? + CD2 + BC? = BD2 + AC2. Therefore the sum of the squares &c. Q E. D.
Cor. From this demonstration, it is manifest that the diameters of every parallelogram bisect one another.
HE radius of a circle is the straight line drawn from the centre
a circle, when it meets the
Straight lines are said to be equally dis
tant from the centre of a circle, when
er perpendicular falls, is said to be
tained by a straight line, and the arch
To find the centre of a given circle.
Draw within it any straight line AB, and bisect (10. 1.) it in D;
For, if it be not, let, if possible, G be the centre, and join GA,
to the less, which is impossible : Therefore G is not the centre of the circle ABC : In the same manner, it can be shown, that no other point but F is the centre : that is, F is the centre of the circle ABC : Which was to be found.
Cor. From this it is manifest that if in a circle a straight line bi. sect another at right angles, the centre of the circle is in the line which bisects the other.
PROP. II. THEOR.
If any two points be taken in the circumference of a circle, the straight
line which joins them shall fall within the circle.
Let ABC be a circle, and A, B any two points in the circumference : the straight line drawn from A to
С B shall fall within the circle.
Take any point in AB as E; find D the centre of the circle ABC ; join AD, DB and DE,and let DE meet the circumference in F. Then, because DA is equal to DB, the an
D gle DAB is equal (5. 1.) to the angle DBA; and because AE, a side of the triangle
E DAE, is produced to B, the angle DEB is
B greater (16. 1.) than the angle DAE ; but
F DAE is equal to the angle DBE; therefore the angle DEB is greater than the angle DBE: Now to the greater angle the greater side is opposite (19. 1.); DB is therefore greater than DE: but BD is equal to DF; wherefore DF is greater than DE, and the point E is therefore within the circle. The same may be demonstrated of any other point between A and B, therefore AB is within the circle. Wherefore, if any two points, &c. Q. E. D.
If a straight line drawn through the centre of a circle bisect a straight
line in the circle, which does not pass through the centre, it will cut that line at right angles ; and if it cut it at right angles, it will bisect it.
· Let ABC be a circle, and let CD, a straight line drawn through the centre, bisect any straight line AB, which does not pass through the centre, in the point F: It cuts it also at right angles.
Take (1. 3.) E the centre of the circle, and join EA, EB. Then because ÀF is equal to FB, and FE common to the two triangles AFE, RFE, there are two sides in the one equal to two sides in the other ;