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“ Cor. 2. From the demonstration it is manifest, that since the

square of CD is quadruple of the square of CB, the square of any - line is quadruple of the square of half that line.”

Otherwise :

** Because AD is divided any how in C (4. 2.), AD-=AC2+CD: - +2CD.AC. But CD=2CB : and therefore CD=CB2+BD2+ * 2CB.BD (4. 2.) = 4CB2, and also 2CD.AC = 4CB.AC ; therefore, " ADP=AC2 +4BC: +4BC.AC. Now BC: +BC.AC=AB.BC(3.2); -- and therefore AD>=AC: +4AB.BC. Q. E. D.”.

PROP. IX. THEOR.

If a straight line be divided into two equal, and also into two unequal parts ; the squares of the two unequal parts are together double of the square of half the line, and of the square of the line between the points of section.

Let the straight line AB be divided at the point C into two equal, and at D into two unequal parts : The squares of AD, DB are together double of the squares of AC, CD.

From the point C draw (11. 1.) CE at right angles to AB, and make it equal to AC or CB, and join EA, EB ; through D draw (31. 1.) DF parallel to CE, and through F draw FG parallel to AB; and join AF : Then, because AC is equal to CE, the angle EAC is equal (5. 1.) to the angle AEC; and because the angle ACE is a right angle, the two others AEC, EAC together make one right angle (32. 1.); and they are equal to one another ; each of them therefore is half of a right angle. For the same reason each of the angles CEP, EBC is half a right angle : and therefore the whole AEB is a right angle : And because the angle GEF is half a right angle, and EGF a right angle, for it is equal (29. 1.) to the interior and opposite angle ECB, A с D B the remaining angle EFG is half a right angle ; therefore the angle GEF is equal to the angle EFG, and the side EG equal (6. 1.) to the side GF: Again, because the angle at B is half a right angle ; and FDB a right angle, for it is equal (29. 1.) to the interior and opposite angle ECB, the remaining angle BFD is half a right angle; therefore the angle at B is equal to the angle BFD, and the side DF to (6. 1.) the side DB. Now, because AC=CE, AC3=CES, and AC2 +CE3=2ACP. But (47. 1.) AE=AC: +CE2 ; therefore AES=24CP. Again, becaụse ÈG=GF, EGPIGF2, and EG+GF2 =2GF2. But EF2=EG? +GP ; therefore, EF2=2GF2 =2CD2, because (34. 1.) CD=GF. And it was shown that AES=2AC2 therefore AES + EF? = 2AC: + 2CDa. But (47. 1.) AF: =AE: +

EF3, and AD? + DF2 = AF, or AD2 + DB2 = AF ; therefore, also AD2 + DB2 = 2AC2 + 2CD2. Therefore, if a straight line, &c. Q. E. D.

Otherwise : “ Because AD = (4. 2.) AC2+CD2+2AC.CD, and DB2+2BC.

CD= (7.2.) BC: +CD=AC2+CD, by adding equals to equals, 56 ADP +BD2 +2BC.CD=2AC2 + 2CD2 +2AC.CD; and therefore

taking away the equal rectangles 2BC.CD and 2AC.CD, there re- mains AD:+DB2=2AC2 +2CD2.”

PROP. X. THEOR. If a straight line be bisected, and produced to any point, the square of the

whole line thus produced, and the square of the part of it produced, are together double of the square of half the line bisected, and of the square of the line made up of the half and the part produced.

Let the straight line AB be bisected in C, and produced to the point D ; the squares of AD, DB are double of the squares of AC, CD.

From the point C draw (11. 1.) CE at right angles to AB, and make it equal to AC or CB ; join AE, EB ; through E draw (31. 1.) EF parallel to AB, and through D draw DF parallel to CE. And because the straight line EF meets the parallels EC, FD, the angles CEF, EFD are equal (29. 1.) to two right angles ; and therefore the angles BEF, EFD are less than two right angles : But straight lines, which with another straight line make the interior angles, upon the same side, less than two right angles, do meet (Cor. 29. 1.) if produced far enough : Therefore EB, FD will meet, if produced towards, B, D ; let them meet in G, and join AG : Then, because AC is equal to CE, the angle CEA is equal (5. 1.) to the angle EAC ; and the angle ACE is a right angle ; therefore each of the angles CEA, EAC is half a right angle (32. 1.): For the same reason, each of the angles CEB, EBC is half à right angle ; therefore AEB is a right angle : And because EBC is half a right angle, DBG is also (15. 1.) half a right angle, for they are vertically opposite; but BDG is a right angle, because it is equal (29.1.) to the alternate angle DCE; therefore the remaining angle DGB is half a right angle, and is therefore equal to the angle DBG; wherefore also the side DB is equal (6. 1.) to the side DG. Again, because EGF is half a right angle, and the angle at F a right angle, being equal (34. 1.) to the opposite angle

E
ÈCD, the remaining angle FEG
is half a right angle, and equal to
the angle EGF ; wherefore also
the side GF is equal (6. 1.) to
the side FE. And because EC

B
CA, EC? + CA2 = 2CAP. A
Now AE2=(47. 1.) AC2 +CES;
therefore, AE? =2AC2. Again,

because EF=FG, EF2=FG', and EF? +FGP=2EF2.' But EGP= (47. 1.) EF2 +, FG? ; therefore EG? = 2EF? ; and since EF = CD, ÈG2=2CD2. And it was demonstrated, that AES=2AC2 ; therefore, AE2 + EG? =2AC2 +2CD2. Now, AGP=AE2 + EG?, wherefore AGP=2AC2+2CD2. But AG(47. 1.) - AD2 +DG2=AD+DB2, because DG=DB: Therefore, AD3 +DBP=2ACP+2CD2. Wherefore, if a straight line, &c. Q. E. D.

PROP. XI. PROB.

To divide a given straight line into two parts, so that the rectangle con

tained by the whole, and one of the parts, may be equal to the square of the other part.

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Let AB be the given straight line ; it is required to divide it into two parts, so that the rectangle contained by the whole, and one of the parts, shall be equal to the square of the other part.

Upon AB describe (46. 1.) the square ABDC ; bisect (10. 1.) AC in E, and join BE ; produce CA to F, and make (3. 1.) ÈF equal to EB, and upon AF describe (46. 1.) the square FGHA ; AB is divided in H, so that the rectangle AB, BH is equal to the square of AH.

Produce GH to K: Because the straight line AC is bisected in E, and produced to the point F, the rectangle CF.FA, together with the square of AE, is equal (6. 2.) to the square of EF : But EF is equal to EB ; therefore the rectangle CF.FA together with the square

of AF, is equal to the square of EB : And the squares of BA, AE are equal (47. 1.) to the square of EB, be

G cause the angle EAB is a right angle ; therefore the rectangle CF.FA, together with the square of AE, is equalto the squares of BA, AE: take away the square of AE, which is common to both, therefore the remaining rectangle CF.FA

HT is equal to the square of AB. Now the figure FK is the rectangle CF.FA, for AF is equal to FG ; and AD is the

square

of AB ; thereforė FK is equal to AD : take away the common part AK, and the remainder FH is equal to the remainder HD. But HD is the rectangle AB.BH, for AB is equal to BD ; and FH is the square of AH ; therefore the rectangle AB.BH is equal to the square of AH :

C

K Wherefore the straight line AB is divided in H so, that the rectangle AB.BH is equal to the square of AH. Which was to be done.

PROP. XII. THEOR.

In obtuse angled triangles, if a perpendicular be drawn from any of

the acute angles to the opposite side produced, the square of the side subtending the obtuse angle is greater than the squares of the sides containing the obtuse angle, by twice the rectangle contained by the side upon which, when produced, the perpendicular falls, and the straight line intercepted between the perpendicular and the obtuse angle.

Let ABC be an obtuse angled triangle, having the obtuse angle 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, BD2 = (4. 2.) BCP+CD+

А 2BC.CD; add AD2 to both: Then BD2 + AD2= BC2 +CD2 + ADS + 2BC.CD. But AB3

- BD2 + AD2 (47. 1.), and AC3=CD? + AD2 (47. 1.); therefore, AB:= BC? + AC2 + 2BC.CD; that is, ABis greater than BC? + AC? by 2BC.CD. Therefore, in ob

В* tuse angled triangles, &c. Q. E. D.

C D

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PROP. XIII. THEOR. In every triangle, the square of the side subtending any of the acute an

gles, 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 perpen. dicular (12. 1.) AD from the opposite angle : The square of AC, opposite to the angle B, is less than the squares of CB, BA by twice the rectangle CB.BD.

First let AD fall within the triangle ABC ; and because the straight line CB is divided into two parts in the point D (7. 2.), BC2 + BD2= 2BC.BD+CD2. Add to each ADP; then BC2+ BD2 +AD2= 2BC.BD +CD2 + AD2, But BD? + AD2

ABP, and CD2 + DA2 = AC2 (47. 1.); therefore BC2 + AB2= 2BC.BD+AC+; that is, AC2 is less

B D than BC2 + ABby 2BC.BD.

Secondly, Let AD fall without the triangle ABC* : Then because the angle at D is a right angle, the angle ACB is greater (16. 1.) than a right angle, and AB2= (12. 2.) AC2+BC2 +2BC.CD. Add BC2 to each ; then AB+-BC2=AC2 + 2BC: + 2BC.CD. But because BD is divided into two parts in C, BC + BC.CD= (3. 2.) BC.BD, and 2BC: + 2BC.CD = 2BC:BD : therefore AB2 + BC2 = AC: + 2BC.BD; and AC? is less than AB2 + BC”, by 2BD.BC.

Lastly, Let the side AC be perpendicular to BC ; then is BC the straight line between the perpendicular and the acute angle at B ; and it is manifest that (47. 1.) AB + BC=AC: +2BC2 =AC +2BC.BC. Therefore in every triangle, &c. Q. E. D.

B

PROP. XIV. PROB.

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To describe a square that shall be equal to a given rectilineal figure.

Let A be the given rectilineal figure ; it is required to describe a .square that shall be equal to A.

Describe (45. 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 semicircle 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 rectangle BE.EF, together with the square of EG, is equal (5. 2.) to the square of GF: but GF is equal to GH ; therefore the rectangle BE.EF, 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 the rectangle BE.EF to

H gether with the square of ÉG is equal to the squares A of HE and EG. Take away the square of EG, which is common to both,

B and the remaining rectangle BE.EF is equal to the square of EH : But

C BD is the rectangle con

* See figure of the last Proposition.

E

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