PROP. 21. THEOR.

If from the ends of one side of a triangle there be drawn two straight lines to a point within the triangle, these two lines shall be less than the two sides of the triangle, but shall contain a greater angle.

1. BA+AE>BE (I. 20), to each add EC;

2.

.. BA+AC>BE+EC.

Again,

CE+ED>CD (I. 20), to each add DB;

.. BE+EC> BD + DC.

But BA+AC has been shown > BE+EC;

.. BA+ AC is still > BD + DC.

:: ▲ BDC > ▲ DEC (I. 16),

and ▲ DEC > ▲ A (I. 16),

.. BDC is still > A. Q.E.D.

51. The three sides of a triangle are together greater than the three medians.

52. The three straight lines drawn from any point within a triangle to the vertices are together less than the perimeter of the triangle.

53. If a trapezium and a triangle have the same base, and the one figure fall within the other, that which has the greater area shall have the greater perimeter. Note.-This proposition is not necessarily true when the trapezium is of the smaller area and has a reflex angle, that is, an angle greater than two right angles.

PROP. 22. PROB.

To make a triangle of which the sides shall be equal to three given straight lines, any two of which are together greater than the third.

Cut off DFA, FGB, GHC (Post. 3).
With F as centre and FD as rad., describe

and with G as centre and GH as rad., describe II.,

cutting I. at K.

F is centre of O I.,

Join KF, KG.

FK FD (Def. 11).

=

But FDA (cons.), .. FK = A.

Similarly, GK GH = C, and FG = B (cons.),

AFGK has its sides respectively equal to A, B, and C. Q.E.F.

54. Show in the above that the circles must cut one another, and that we can make two triangles satisfying the given conditions.

55. Show by diagrams what form the figure would take if one of the given lines be taken (1) greater than, (2) equal to, the other two together.

56. Construct a rhombus on a given line, one of the diagonals of which shall be equal to the given line.

57. Tstruct a right-angled triangle, (1) with one side nuse equal respectively to two given lines; ual respectively to two given lines.

PROP. 23. PROB.

At a given point in a given straight line, to make an angle equal to a given angle, i.e., so that the given line shall be one of the arms of the angle.

E

A A

D

A

Given line AB with point A in it, and ≤ C.
To make at A in AB an angle

=

< C.

B

Take D and E, any points in CD, CE, and join them. Cut off from AB, AF CD (Post. 3), and from A as centre, with CE as rad., describe a ; from Fas centre, with DE as rad., describe a O, cutting the first at G (Post. 3). Join AG and FG.

=

AF .. AFAG

=

CE, and FG DE (cons.),

=

CD, AG
ADCE (1. 8). .. LA = 4 C.

=

.. at A in AB an has been described

= ▲ C.

Q.E.F.

Note. In this proposition, after joining DE, Euclid says, "Make a triangle AFG whose sides shall be equal to CD, DE, and EC (I. 22)," and then proof follows as above by I. 8. It seems better to make our construction more detailed, following the method of I. 22, that we may be certain of the position of the constructed triangle.

58. To make a triangle whose sides shall be respectively equal to the halves of the sides of a given triangle; or to the doubles of the sides of a given triangle.

59. To construct a triangle having given the base and the two angles at the base, which are together less than two right angles (I. 17).

60. To construct a triangle whose base shall be equal to a given line, one of the angles at the base equal to a given angle, and the sum of the other two sides equal to a given line.

PROP. 22. PROB.

To make a triangle of which the sides shall be equal to three given straight lines, any two of which are together greater than the third.

Cut off DF A, FGB, GH = C (Post. 3).

=

With F as centre and FD as rad., describe

I.;

and with G as centre and GH as rad., describe II.,

F is centre of I., FK = FD (Def. 11).

But FDA (cons.), . FK = A.

Similarly, GK = GH = C, and FG = B (cons.),

AFGK has its sides respectively equal to A,
and C. Q.E.F.

B,

54. Show in the above that the circles must cut one another, and that we can make two triangles satisfying the given conditions.

55. Show by diagrams what form the figure would take if one of the given lines be taken (1) greater than, (2) equal to, the other two together.

56. Construct a rhombus on a given line, one of the diagonals of which shall be equal to the given line.

uct a right-angled triangle, (1) with one side use equal respectively to two given lines; ual respectively to two given lines.

PROP. 23. PROB.

At a given point in a given straight line, to make an angle equal to a given angle, i.e., so that the given line shall be one of the arms of the angle.

Given line AB with point A in it, and ▲ C.
To make at A in AB an angle

=

=

< C.

B

Take D and E, any points in CD, CE, and join them. Cut off from AB, AF CD (Post. 3), and from A as centre, with CE as rad., describe a O; from F as centre, with DE as rad., describe a O, cutting the at G (Post. 3). Join AG and FG.

first :: AF

=

==

CD, AG = CE, and FG = DE (cons.),

.. AFAG ADCE (1. 8).

=

:: LA = 4 C.

.. at A in AB an has been described

=

▲ C.

Q.E.F.

Note. In this proposition, after joining DE, Euclid says, "Make a triangle AFG whose sides shall be equal to CD, DE, and EC (I. 22)," and then proof follows as above by I. 8. It seems better to make our construction more detailed, following the method of I. 22, that we may be certain of the position of the constructed triangle.

58. To make a triangle whose sides shall be respectively equal to the halves of the sides of a given triangle; or to the doubles of the sides of a given triangle.

59. To construct a triangle having given the base and the two angles at the base, which are together less than two right angles (I. 17).

60. To construct a triangle whose base shall be equal to a given line, one of the angles at the base equal to a given angle, and the sum of the other two sides equal to a given line.