PROPOSITION I. PROBLEM.

To trisect a given straight line.

ANALYSIS. Let AB be the given straight_line, and suppose it divided into three equal parts in the points D, E.

On DE describe an equilateral triangle DEF,
then DF is equal to AD, and FE to EB.
On AB describe an equilateral triangle ABC,
and join AF, FB.

Then because AD is equal to DF,

therefore the angle AFD is equal to the angle DAF, and the two angles DAF, DFA are double of one of them DAF. But the angle FDE is equal to the angles DAF, DFA, and the angle FDE is equal to DAC, each being an angle of an equilateral triangle;

therefore the angle DAC is double the angle DAF;

wherefore the angle DAC is bisected by AF.

Also because the angle FAC is equal to the angle FAD,
and the angle FAD to DFA;

therefore the angle CAF is equal to the alternate angle AFD: and consequently FD is parallel to AC.

Synthesis. Upon AB describe an equilateral triangle ABC, bisect the angles at A and B by the straight lines AF, BF, meeting in F; through F draw FD parallel to AC, and FE parallel to BC. Then AB is trisected in the points D, E.

For since AC is parallel to FD and FA meets them,
therefore the alternate angles FAC, AFD are equal;
but the angle FAD is equal to the angle FAC,
hence the angle DAF is equal to the angle AFD,
and therefore DF is equal to DA.

But the angle FDE is equal to the angle CAB,
and FED to CBA; (1. 29.)

therefore the remaining angle DFE is equal to the remaining angle АСВ.

Hence the three sides of the triangle DFE are equal to one another, and DF has been shewn to be equal to DA,

therefore AD, DE, EB are equal to one another.

Hence the following theorem.

If the angles at the base of an equilateral triangle be bisected by two lines which meet at a point within the triangle; the two lines drawn from this point parallel to the sides of the triangle, divide the base into three equal parts.

Note. There is another method whereby a line may be divided into three equal parts :-by drawing from one extremity of the given line, another making an acute angle with it, and taking three equal distances from the extremity, then joining the extremities, and through the other two points of division, drawing lines parallel to this line through the other two points of division, and to the given line; the three triangles thus formed are equal in all respects. This may be extended for any number of parts, and is a particular case of Euc. vi. 10.

PROPOSITION II. THEOREM.

If two opposite sides of a parallelogram be bisected, and two lines be drawn from the points of bisection to the opposite angles, these two lines trisect the diagonal.

Let ABCD be a parallelogram of which the diagonal is AC.
Let AB be bisected in E, and DC in F,

also let DE, FB be joined cutting the diagonal in G, II.
Then AC is trisected in the points G, H.

Through E draw EK parallel to AC and meeting FB in K. Then because EB is the half of AB, and DF the half of DC; therefore EB is equal to DF;

and these equal and parallel straight lines are joined towards the same parts by DE and FB;

therefore DE and FB are equal and parallel. (1. 33.)

And because AEB meets the parallels EK, AC,

therefore the exterior angle BEK is equal to the interior angle EAG. For a similar reason, the angle EBK is equal to the angle AEG. Hence in the triangles AEG, EBK, there are the two angles GAE, AEG in the one, equal to the two angles KEB, EBK in the other, and one side adjacent to the equal angles in each triangle, namely AE equal to EB;

therefore AG is equal to EK, (1. 26.) but EK is equal to GH, (1. 34.) therefore AG is equal to GH. By a similar process, it may be shewn that GH is equal to HC. Hence AG, GH, HC are equal to one another.

It

and therefore AC is trisected in the points G, H.

may also be proved that BF is trisected in H and K.

PROPOSITION III. PROBLEM.

Draw through a given point, between two straight lines not parallel, a straight line which shall be bisected in that point.

Analysis. Let BC, BD be the two lines meeting in B, and let A be the given point between them.

Suppose the line EAF' drawn through A, so that EA is equal to AF;

through A draw AG parallel to BC, and GH parallel to EF. Then AGHE is a parallelogram, wherefore AE is equal to GH, but EA is equal to AF by hypothesis; therefore GH is equal to AF. Hence in the triangles BIG, GAF,

the angles HBG, AGF are equal, as also BGH, GFA, (1. 29.)
also the side GH is equal to AF;

whence the other parts of the triangles are equal, (1. 26.)
therefore BG is equal to GF.

Synthesis. Through the given point A, draw AG parallel to BC on GD, take GF equal to GB;

then Fis a second point in the required line:
join the points F, A, and produce FA to meet BC in E;
then the line FE is bisected in the point A;
draw GH parallel to AE.

Then in the triangles BGH, GFA, the side BG is equal to GF and the angles GBH, BGH are respectively equal to FGA, GFA; wherefore GH is equal to AF, (1. 26.) but GH is equal to AE, (1. 34.)

therefore AE is equal to AF, or EF is bisected in 4.

PROPOSITION IV. PROBLEM.

From two given points on the same side of a straight line given in position, draw two straight lines which shall meet in that line, and make equal angles with it; also prove, that the sum of these two lines is less than the sum of any other two lines drawn to any other point in the line.

Analysis. Let A, B be the two given points, and CD the given line. Suppose G the required point in the line, such that AG and BG being joined, the angle AGC is equal to the angle BGD.

Draw AF perpendicular to CD and meeting BG produced in E. Then, because the angle BGD is equal to AGF, (hyp.) and also to the vertical angle FGE, (1. 15.)

therefore the angle AGF is equal to the angle EGF;

also the right angle AFG is equal to the right angle EFG, and the side FG is common to the two triangles AFG, EFG, therefore AG is equal to EG, and AF to FE.

Hence the point E being known, the point G is determined by the intersection of CD and BE.

Synthesis. From A draw AF perpendicular to CD, and produce it to E, making FE equal to AF, and join BE cutting CD in G. Join also AG.

Then AG and BG make equal angles with CD.

For since AF is equal to FE, and FG is common to the two triangles AGF, EGF, and the included angles AFG, EFG are equal; therefore the base AG is equal to the base EG,

and the angle AGF to the angle EGF;

but the angle EGF is equal to the vertical angle BGD,
therefore the angle AGF is equal to the angle BGD;

that is, the straight lines AG and BG make equal angles with the straight line CD.

Also the sum of the lines AG, GB is a minimum.

For take any other point H in CD, and join EH, HB, AH. Then since any two sides of a triangle are greater than the third side, therefore EH, HB are greater than EB in the triangle EHB. But EG is equal to AG, and EH to AH;

therefore AH, HB are greater than AG, GB.

That is, AG, GB are less than any other two lines which can be drawn from A, B, to any other point H in the line CD.

By means of this Proposition may be found the shortest path from one given point to another, subject to the condition, that it shall meet two given lines.

PROPOSITION V. PROBLEM.

Given one angle, a side opposite to it, and the sum of the other two sides, construct the triangle.

Analysis. Suppose BAC the triangle required, having BC equal to the given side, BAC equal to the given angle opposite to BC, also BD equal to the sum of the other two sides.

D

B

Join DC.

Then since the two sides BA, AC are equal to BD, by taking BA from these equals, the remainder AC is equal to the remainder AD. Hence the triangle ACD is isosceles, and therefore the angle ADC is equal to the angle ACD.

But the exterior angle BAC of the triangle ADC is equal to the two interior and opposite angles ACD and ADC:

Wherefore the angle BAC is double the angle BDC, and BDC is the half of the angle BAC.

Hence the synthesis.

At the point D in BD, make the angle BDC equal to half the given angle,

and from B the other extremity of BD, draw BC equal to the given side, and meeting DC in C,

at C in CD make the angle DCA equal to the angle CDA, so that CA may meet BD in the point A.

Then the triangle ABC shall have the required conditions.

PROPOSITION VI. PROBLEM.

To bisect a triangle by a line drawn from a given point in one of the sides. Analysis. Let ABC be the given triangle, and D the given point in the side AB.

Suppose DF the line drawn from D which bisects the triangle; therefore the triangle DBF is half of the triangle ABC. Bisect BC in E, and join AE, DE, AF,

then the triangle ABE is half of the triangle ABC: hence the triangle ABE is equal to the triangle DBF; take away from these equals the triangle DBE, therefore the remainder ADE is equal to the remainder DEF. But ADE, DEF are equal triangles upon the same base DE, and on the same side of it,

they are therefore between the same parallels, (I. 39.)
that is, AF is parallel to DE,
therefore the point Fis determined.
Synthesis. Bisect the base BC in E, join DE,
from A, draw AF parallel to DE, and join DF.
Then because DE is parallel to AF,

therefore the triangle ADE is equal to the triangle DEF; (1.37.)
to each of these equals, add the triangle BDE,
therefore the whole triangle ABE is equal to the whole DBF,
but ABE is half of the whole triangle ABC;
therefore DBF is also half of the triangle ABC.

PROPOSITION VII. THEOREM.

If from a point without a parallelogram lines be drawn to the extremities of two adjacent sides, and of the diagonal which they include; of the triangles thus formed, that, whose base is the diagonal, is equal to the sum of the other two.

Let ABCD be a parallelogram of which AC is one of the diagonals, and let P be any point without it: and let AP, PC, BP, PD be joined.

Then the triangles APD, APB are together equivalent to the tri angle APC.

B