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If a straight line be bisected, and produced to any point, the square on the whole line thus produced, and the square on the part of it produced, are together double of the square on half the line bisected, and of the square on the line made up of the half and the part produced.

Given the straight line AB, bisected in C, and produced to the point D; to prove that the squares on AD and DB are double

of the squares on AC and CD.

(Const.) From the point C draw CE at right angles to AB and make it equal to AC or CB,

(1. 11);

EB;

through E draw EF parallel to

and through D draw DF

And because the

AB (I. 31),
parallel to CE.
straight line EF meets the parallels EC,
FD, the angles CEF and EFD are
together equal to two right angles (I. 29);

and therefore the angles BEF, EFD

are less than two right angles.

But

straight lines, which with another straight

and join AE,

F

B

G

line make the interior angles upon the same side less than two right angles, do meet if produced far enough (I. 29, Cor.); therefore EB and FD shall meet, if produced towards B and D ; let them meet in G, and join AG. (Dem.) Then, because AC is equal to CE, the angle CEA is equal to the angle EAC (I. 5); and the angle ACE is a right angle; therefore each of the angles CEA, EAC is half a right angle (I. 32). For the same reason, each of the angles CEB and EBC is half a right angle; therefore the angle AEB is a right angle.

And because EBC is half a right angle, DBG is also half a right angle (I. 15), for they are vertically opposite; but BDG is a right angle, because it is equal to the alternate angle DCE (I. 29); therefore the remaining angle DGB is half a right angle (I. 32), and is therefore equal to the angle DBG; wherefore also the side DB is equal to the side DG (I. 6).

Again, because EGF is half a right angle, and that the angle at F is a right angle, because it is equal to the opposite angle ECD (I. 34), the remaining angle FEG is half a right angle, and equal to the angle EGF;

wherefore also the side GF

is equal to the side FE. And because EC is equal to CA, the square on EC is equal to the square on CA; therefore the squares on EC and CA are double of the square on CA. But the square on EA is equal to the squares on EC and CA (I. 47); therefore the square on EA is double of the square on Again, because GF is equal to FE, the square on GF is equal to the square on FE; and therefore the squares on GF and FE are double of the square on EF. But the square

AC.

on EG is equal to the squares on GF and FE (I. 47); fore the square on EG is double of the square on EF; is equal to CD;

square on CD;

thereand EF

wherefore the square on EG is double of the but it was demonstrated that the square on EA is double of the square on AC; therefore the squares on AE and EG are double of the squares on AC and CD. And the square on AG is equal to the squares on AE and EG; therefore the square on AG is double of the squares on AC and CD. But the squares on AD and DG are equal to the square on AG; therefore the squares on AD and DG are double of the squares on AC and CD. But DG is equal to DB; therefore the squares on AD and DB are double of the squares on AC and CD.

Otherwise: The same as the 'Otherwise' given in last proposition.

Schol. The last two propositions are equivalent to the following: If a straight line be bisected, and be also cut unequally, either internally or externally, the sum of the squares on the two unequal segments is equal to twice the squares on half the line and on the mean distance.

PROPOSITION XI. PROBLEM.

To divide a given straight line in medial section.

Given a straight line AB; it is required to divide it in medial section.

(Const.) Upon AB describe the square ACDB (I. 46);

bisect

F

G

AC in E (I. 10), and join BE; produce CA to F, and make EF equal to EB (I. 3), and upon AF describe the square FAHG, and produce GH to K; AB is divided in H, so that the rectangle AB BH is equal to the square of AH.

(Dem.) Because the straight line AC is bisected in E, and produced to the point F,

the rectangle CF FA, together with the square on AE, is equal to the square on EF (II. 6). But EF is equal to EB; therefore the rectangle CF FA, together with the square on AE, is equal to the square on EB;

A

E

!H

B

C

K

and the squares on BA and AE are equal to the square on EB (I. 47), because the angle EAB is a right angle; therefore the rectangle CF FA, together with the square on AE, is equal to the squares on BA and AE; take away the square on AE, which is common to both; therefore the remaining rectangle CF FA is equal to the square on AB.* Now the figure FK is the rectangle contained by CF and FA, for AF is equal to FG, and AD is the square on AB; therefore FK is equal to AD. Take away the common part AK, and the remainder FH is equal to the remainder

HD.

AB is equal to BD,

And HD is the rectangle contained by AB and BH, for and FH is the square on AH; therefore the rectangle AB BH is equal to the square on AH. Wherefore the straight line AB is divided in H, so that the rectangle AB BH is equal to the square on AH.

*COR.-The line CF is also divided medially in A; but AC is equal to AB and AF to AH; hence, the line made up of the whole and its greater segment is also divided in medial section.

PROPOSITION XII. THEOREM.

In obtuse-angled triangles, if a perpendicular be drawn from any of the acute angles to the opposite side produced, the square on the side subtending the obtuse angle is greater than the squares on 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 without the triangle between the perpendicular and the obtuse angle.

T

Given an obtuse-angled triangle ABC, having the obtuse angle ACB, and from the point A let AD be drawn perpendicular to BC produced (I. 12); to prove that the square on AB is greater than the squares on AC and CB, by twice the rectangle BC CD.

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(Dem.) Because the straight line BD is divided into two parts in the point C, the square on BD is equal to the squares on BC and CD, and twice the rectangle BCCD (II. 4); to each of these equals add the square on DA; and the squares on DB and DA are equal to the squares on BC, CD, and DA, and twice the rectangle BC CD. But the square on BA is equal to the squares on BD and DA (I. 47), because the angle at D is a right angle; square on CA is equal to the squares on CD and DA. fore the square on BA is equal to the squares on BC and CA, and twice the rectangle BC CD; that is, the square on BA is greater than the squares on BC and CA, by twice the rectangle BC CD.

and the

There

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In every triangle, the square on the side subtending any of the acute angles is less than the squares on 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.

Given any triangle ABC, and the angle at B one of its acute angles, and upon BC, one of the sides containing it, let fall the

perpendicular AD from the opposite angle (I. 12); to prove that the square on AC, opposite to the angle B, is less than

the squares on CB and BA, by twice the rectangle CB BD.

(Dem.) First, Let AD fall within the triangle ABC; and because the straight line CB is divided into two parts in the point D, the squares on CB and BD are

B

and the

equal to twice the rectangle contained by CB, BD, and the square on DC (II. 7). To each of these equals add the square on AD; therefore the squares on CB, BD, and DA are equal to twice the rectangle CB BD, and the squares on AD and DC. But the square on AB is equal to the squares on BD and DA (I. 47), because the angle BDA is a right angle; square on AC is equal to the squares on AD and DC. fore the squares on CB and BA are equal to the square on AC, and twice the rectangle CB BD; that is, the square on AC alone is less than the squares on CB and BA, by twice the rectangle CB⚫BD.

There

then,

Secondly, Let AD fall without the triangle ABC; because the angle at D is a right angle, the angle ACB is greater than a right angle (I. 16); and therefore

the square on AB is equal to the squares on
AC and CB, and twice the rectangle BC CD
(II. 12). To these equals add the square on
BC;
therefore the squares on AB and
BC are equal to the square on AC, and twice
the square on BC, and twice the rectangle
BC CD. But because BD is divided into
two parts in C,

BC CD, and the

the rectangle DB BC is equal to the rectangle square on BC (II. 3); and the doubles of

these are equal; therefore the squares on AB and BC are equal to the square on AC, and twice the rectangle

DB⚫ BC;

therefore the square on AC alone is less than the squares on AB and BC, by twice the rectangle DB 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 the squares on AB and BC are equal to the square on AC, and twice the square on BC (I. 47).

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A

In any triangle, if a perpendicular be let fall from the vertex upon the base, then, according as the angle opposite to one of the

sides is obtuse or acute, the square on that side is greater or less than the squares on the base and the other side, by twice the rectangle contained by the base and the part of it intercepted between that angle and the perpendicular.

Given any triangle ABC, and CD a perpendicular from C upon AB or AB produced; to prove that AC2 = AB2 + BC2± taking the upper sign when the angle ABC is and the lower when it is acute.

2AB⚫ BD, obtuse,

АДА

B D

D B

A

For AD2 = AB2 + BD2 ± 2AB· BD (II. 4, and 7 Cor. 2),

add DC2 to both,

then AD2 + DC2 = AB2 + DB2 +DC2 ± 2AB BD.

But (I. 47) AD2 + DC2 = AC2,

and BD2+ DC2 = BC2,

therefore AC2 = AB2 + BC2± 2AB BD.

In the demonstration + refers to the first diagram, and to the second and third.

Schol. If the sides opposite the angles A, B, and C be called a, b, and c, the result becomes,

b2 = c2 + a2 ± 2c × BD,

from which we obtain in the first diagram, where the angle B is obtuse,

b2 BD =

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and from the second and third diagrams, where the angle B is acute,

a2 + c2 62

BD =

2c

From these formulæ, the segments of the sides made by a perpendicular from any of the angles may readily be calculated when the three sides are known.

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

Given the rectilineal figure A; square that shall be equal to A.

it is required to describe a

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