The First Six Books of the Elements of Euclid, with a Commentary and Geometrical Exercises

(72) From a given point (C) in a given right line (AB) to draw a perpendicular to the given line.

Solution.

In the given line take any point D and make C E equal to CD (III), upon D E describe an equilateral triangle D FE (I), draw F C, and it is perpendicular to the given line.

Demonstration.

Because the sides DF and DC are equal to the sides EF and E C (const.), and CF is common to the triangles D F C and E FC, therefore (VIII) the angles opposite to the equal sides D F and EF are equal, and therefore FC is perpendicular to the given right line A B at the point C. COR.-By help of this problem it may be demonstrated, that two straight lines cannot have a common segment.

If it be possible, let the two straight lines A B C, A B D have the segment A B common to both of them. From the point B draw BE at right angles to A B; and because A B ̊C is a straight line, the angle CBE is equal to the

angle EBA; in the same manner, because ABD is a straight line, the angle DBE is equal to the angle E BA; wherefore the angle

D'BE is equal to the angle CBE, the less to the greater, which is impossible; therefore two straight lines cannot have a common segment.

If the given point be at the extremity of the given right line, it must be produced, in order to draw the perpendicular by this construction.

In a succeeding article, the student will find a method of drawing a perpendicular through the extremity of a line without producing it. The corollary to this proposition is useless, and is omitted in some editions.

It is equivalent to proving that a right line cannot be produced through its extremity in more than one direction, or that it has but one production.

(73) To draw a perpendicular to a given indefinite right line (AB) from a point (C) given without it.

Solution.

Take any point X on the other side of the given line, and from

5° The centre of the second circle is that vertex of the triangle which is opposite to the joining line, and its radius is made up of that side of the triangle which is opposite to the given point, and its production which is the radius of the first circle. So that the radius of the second circle is the sum of the side of the triangle and the radius of the first circle.

6° The side of the equilateral triangle which is produced through the given point to meet the second circle, is that side which is opposite to the connected extremity of the given right line, and the production of this side is the line which solves the problem; for the sum of this line and the side of the triangle is the radius of the second circle, but also the sum of the given right line (which is the radius of the first circle) and the side of the triangle is equal to the radius of the second circle. The side of the triangle being taken away the remainders are equal.

As the given point may be joined with either extremity, there may be two different joining lines, and as the triangle may be constructed on either side of each of these, there may be four different triangles; so the right line and point being given, there are four different constructions by which the problem may be solved.

If the student inquires further, he will perceive that the solution may be effected also by producing the side of the triangle opposite the given point, not through the extremity of the right line but through the vertex of the triangle. The various consequences of this variety in the construction we leave to the student to trace. (60) By the second proposition a right line of a given length can be inflected from a given point P upon any given line A B. For from the point P draw a right line of the given length (II), and with P as centre, and that line as radius, describe a circle. A line drawn from P to any point C, where this circle meets the given line A B, will solve the problem.

By this proposition the first may be generalized; for an isosceles triangle may be constructed on a given line as base, and having its side of a given length. The construction will remain unaltered, except that the radius of each of the circles will be equal to the length of the side of the proposed triangle. If this length be not greater than half the base, the two circles will not intersect, and no triangle can be constructed, as will appear hereafter.

(61) From the greater (A B), of two given right lines to cut off a part equal to the less (C).

Solution.

From either extremity A of the greater let a right line A D be drawn equal to the less C (II), and with the point A as centre, and the radius AD let a circle be described (41). The part A E of the greater cut off by this circle will be equal to the less C.

Demonstration.

For A E and AD are equal, being radii of the same circle (17); and C and A D are equal by the construction. Hence A E and C are equal.

By a similar construction, the less might be produced until it equal the greater. From an extremity of the less let a line equal to the greater be drawn, and a circle be described with this line as radius. Let the less be produced to meet this circle.

(62) If two triangles (BAC and EDF) have two sides (B A and A C) in the one respectively equal to two sides (E D and DF) in the other, and the angles (A and D) included by those sides also equal; the bases or remaining sides (B C and EF) will be equal, also the angles (B and C) at the base of the one will be respectively equal to those (E and F) at the base of the other which are opposed to the equal sides (i. e. B to E and C to F).

Let the two triangles be conceived to be so placed that the vertex of one of the equal angles D shall fall upon that of the other A, that one of the sides DE containing the given equal angles shall fall upon the side A B in the other triangle to which it is equal, and that the remaining pair of equal sides AC and D F shall lie at the same side of those AB and D E which coincide.

Since then the vertices A and D coincide, and also the equal sides A B and D E, the points B and E must coincide. (If they did not the sides A B and D E would not be equal.) Also, since the side D E falls on A B, and the sides A C and D F are at the same side of A B, and the angles A and D are equal, the side D F must fall upon A C; (for otherwise the angles A and D would not be equal.)

Since the side D F falls on A C, and they are equal, the extremity F must fall on C. Since the extremities of the bases B C and EF coincide, these lines themselves must coincide, for if they did not they would include a space (52). Hence the sides B C and E F are equal (50).

Also, since the sides E D and E F coincide respectively with B A and B C, the angles E and B are equal (50), and for a similar reason the angles F and C are equal.

Since the three sides of the one triangle coincide respectively with the three sides of the other, the triangles themselves coincide, and are therefore equal (50).

In the demonstration of this proposition, the converse of the eighth axiom (50) is assumed. The axiom states, that if two magnitudes coincide they must be equal.' In the proposition it is assumed, that if they be equal they must under certain circumstances coincide. For when the point D is placed on A, and the side D E on A B, it is assumed that the point E must fall upon B, because A B and D E are equal. This may, however, be proved by the combination of the eighth and ninth axioms; for if the point E did not fall upon B, but fell either above or below it, we should have either E D equal to a part of B A, or B A equal to a part of E D. In either case the ninth axiom would be contradicted, as we should have the whole equal to its part.

The same principle may be applied in proving that the side DF will fall upon A C, which is assumed in Euclid's proof.

In the superposition of the triangles in this proposition, three things are to be attended to:

1o The vertices of the equal angles are to be placed one on the other.

2o Two equal sides to be placed one on the other.

3o The other two equal sides are to be placed on the same side of those which are laid one upon the other.

From this arrangement the coincidence of the triangles is inferred. It should be observed, that this superposition is not assumed to be actually effected, for that would require other postulates besides the three already stated; but it is sufficient for the validity of the reasoning, if it be conceived to be possible that the triangles might be so placed. By the same principle of superposition, the following theorem may be easily demonstrated,' If two triangles have two angles in one respectively equal to two angles in the other, and the sides lying between those angles also equal, the remaining sides and angles will be equal, and also the triangles themselves will be equal.' prop. xxvi.

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This being the first theorem in the Elements, it is necessarily deduced exclusively from the axioms, as the first problem must be from the postulates. Subsequent theorems and problems will be deduced from those previously established.

PROPOSITION V. THEOREM.

(63) The angles (B, C) opposed to the equal sides (A C and A B) of an isosceles triangle are equal, and if the equal sides be produced through the extremities (B and C) of the third side, the angles (D B C and E C B) formed by their produced parts and the third side are equal.

Let the equal sides A B and A C be produced through the extremities B, C of the third side, and in the produced part B D of either let any point F be assumed, and from the other let A G be cut off equal to AF (III). Let the points F and G so taken on the produced sides be connected by right lines F C and B G with the alternate extremities of the third side of the triangle.

In the triangles F A C and G AB the sides FA and A C are respectively equal to G A and A B, and the included angle A is common to both triangles. Hence (IV), the line FC is equal to B G, the angle AFC to the angle A G B, and the angle A C F to the angle A B G. If from the equal lines A F and A G, the equal sides A B and A C be taken, the remainders BF and C G will be equal. Hence, in the triangles B F C and C G B, the sides B F and F C are respectively equal to C G and G B, and the angles F and G included by those sides are also equal. Hence (IV), the angles FBC and G C B, which are those included by the third side B C and the productions of the equal sides A B and A C, are equal. Also, the angles F C B and G B C are equal. If these equals be taken from the angles F CA and G B A, before proved equal, the remainders, which are the angles A B C and A C B opposed to the equal sides, will be equal.

(64) COR.-Hence, in an equilateral triangle the three angles are equal; for by this proposition the angles opposed to every two equal sides are equal.

(65) If two angles (B and C) of a triangle (B A C) be equal, the sides (A C and A B) opposed to them are also equal.

For if the sides be not equal, let one of them A B be greater than the other, and from it cut off D B equal to A C (III), and draw C D.

Then in the triangles DB C and ACB, the sides DB and B C are equal to the sides A C and CB respectively, and the angles D B C and AC B are also equal; therefore (IV) the triangles themselves DBC and ACB are equal, a part equal to the whole, which is absurd; therefore neither of the sides A B or A C is greater than the other; they are therefore equal to one another.

(66) COR.-Hence every equiangular triangle is also equilateral, for the sides opposed to every two equal angles are equal.

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