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ACB, and from the point A let AD be drawn (12. 1.) perpendicular to BC produced: The square of AB is greater than the squares of AC, CB, by twice the rectangle BC.CD.
Because the straight line BD is divided into two parts in the point C, BD= (4. 2.) BCS + CDs +
А. 2BC.CD; add AD' to both: Then BD? + AD: = BC+ CD +AD% + 2BC.CD. But AB' =BD? + AD (47. 1.), and AC'=CD+ AD? (49. 1.); therefore, AB'S BC? + AC: + 2BC.CD; that is, AB is greater than BC +AC by 2BC.CD. Therefore, in abtuse angled triangles, &c. Q.E.D. B
PROP. XIII, THEOR, In every triangle the square of the side subtending any of the acute angles, is less than the squares of the sides containing that angle, by twice the rectangle contained by either of these sides, and the straight line intercepted between the perpendicular, let fall upon it from the opposite angle, and the acute angle.
Let ABC be any triangle, and the angle at B one of its acute angles, and upon BC, one of the sides containing it, let fall the perpendicular (12. 1.) AD from the opposite angle: The square of AC, opposite to the angle B, is less than the squares af CB, BA by twice the rectangle CB.BD. First let AD fall within the trian:
А gle ABC; and because the straight line CB is divided into two parts in the point D (7. 2.), BC +BD'= 2BC.BD+CD. Add to each ADs; then BC + BD + AD = 2BC.BD +CD% + AD%. But BD' + AD”
AB%, and CDs + DA” AC (47. 1.); therefore BC + AB:= 2BC.BD+AC'; that is, AC is less
11 than BC + AB. by 2BC.BD. + Secondly, Let AD fall without the triangle ABC*: Then because the angle at D is a right angle, the angle AČB is greater (16. 1.) than a right angle, and AB'=(12. 2.) ACS+BC°+2BC.CD. Add BC to each; then AB'+ BC =ĄC2+2BC: +2BC.CD. But because *BD is divided into two parts in C, BCP+BC.CD= (3. 2.) BC.BD, and 2BC% + 2BC.CD 2BO.BD: therefore ABS + BC=AC+ 2BC.BD; and AC is less than AB'+BC, by 2BD.BC.
See figure of the last Proposition.
Lastly, Let the side AC be perpen. dicular to BC; then is BC the straight line between the perpendicular and the acute angle at B; and it is manifest that (47. 1.) ABS + BC = AC"+2BC:=AC +2BC.BC. Therefore in every triangle, &c. Q. E. D.
PROP. XIV. PROB.
To describe a square that shall be equal to a given rectili
neal figure. Let A be the given rectilineal figure; it is required to describe a square that shall be equal to A.
Describe (48.1.) the rectangular parallelogram BCDE equal to the rectilineal figure A. If then the sides of it, BE, ED are equal to one another, it is a square, and what was required is done; but if they are not equal, produce one of them, BE to F, and make EF equal to ED, and bisect BF in G: and from the centre G, at the distance GB, or GF, describe the semieitele BHF and produce DE to H, and join GH. Therefore, because the straight line BF is divided into two equal parts in the point G, and into two unequal in the point E, the reetangle BE.EF, together with the square of EG, is equal (6. 2.) to the square of GF: bat GF is equal to GH; therefore the rectangle BEEF, together with the square of EG, is equal to the square of GH: But the squares of HE and EG are equal (47. 1.) to the square of GH: Therefore also
H the rectangle BE.EF together with the
of EG, is equal to the squares of HE and EG. Take away
square which is common to both, and the remaining rect
E angle BE.EF is equal to the square of EH: But
D BD is the rectangle contained 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 bas been made equal to the given rectilineal tigare A, viz. the square described upen 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 A E perpendicular to BC, and because BEA is a right angle, AB = (47. 1.) BE +AE and AC = CES + AE%; wherefore AB+ AC% BE: + CE2 + 2AE%. But because the line BC is cut equally in D, and unequally in E, BE + ČE = (9. 2.) 2BD + 2DE; therefore AB'+ AC2BD2+2DE. 2AEP.
Now DE? + AE = (47. 1.) AD, and 2DE2 + 2AE2 2AD2; wherefore ABP +AC2=2BD2 +2AD, There.B.
D E C fore, &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 (18. 1.) and also the alternate angles EAD, ECB (29. 1.), the triangles ADE, CEB hate 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.
E 1.); therefore the other sides which are opposite to the equal angles are also equal (26. 1.), viz. AE to EC, and ED to EB.
B Since, therefore, BD is bisected in E, AB'+AD'=(A. 2.) 2BE. +2AE”; and for the same reason, CD+BCʻ=2BE2 +2EC=2BE: +2AE, because EC=AE. Therefore AB+AD+DC +BC'= 4BE +4AE". But 4BE’=BD, and 4AE'=AC (2. Cor. 8. 2.) because BD and AC are both bisected in E; therefore AB + A027 CD + BC =BD +AC. 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.
to the circumference.
a circle, when it meets the
another, which meet, but do
tant from the centre of a circle, when
er perpendicular falls, is said to be
tained by a straight line, and the arch
ed by two straight lines drawn from any
the arch intercepted between the straight
tained by two straight lines drawn from
Similar segments of a circle,
are those in which the angles are equal, or which contain equal angles.
PROP. I. PROB.
Draw within it any straight line AB, and bisect (10. 1.) it in D; from the point D draw (11. 1.) DC at right angles to AB, and produce it to E, and bisect CE in F : the point F is the centre of the circle ABC.
For, if it be not, let, if possible, G be the centre, and join GA, GD, GB: Then, because DA is equal to DB, and DG common to the two triangles ADG, BDG, the two sides AD, DG are equal to the two BD, DG, each to each; and the base GA is equal to the base GB, because they are radii of the same circle: therefore the angle ADG is equal (8. 1.) to the angle GÖB: But when a straight line standing up
G on another straight line makes the adjacent F angles equal to one another, each of the angles is a right angle (7. def. 1.) Therefore
A А the angle GDB is a right angle: But FDB is
D likewise a right angle; wherefore the angle FDB is equal to the angle GDB, the greater