PROP. XXI. fig. 34. If two triangles, as ABC and DEF, have any angle of the one equal to an angle of the other, as the angles at B and E, and the sides forming these angles respectively proportional, that is, AB: BC:: DE: EF; then these triangles will be similar to each other. In the greater triangle DEF, make EH equal to BC of the less triangle, and draw HG parallel to the side DF, then will the angle EHG be equal to the angle EFD, and the triangle EGH be equiangular to the triangle EFD; and consequently EF: EH:: ED: EG: but by the proposition EF: BC:: ED: BA, and EH being made equal to BC, EG will be equal to BA, and the triangles EHG and BCA will be equal: but the triangle EHG was already shown to be similar to EFD; therefore the triangle ABC will be similar to the triangle DEF.

From these properties of similar triangles it may be demonstrated, that the areas of such triangles are to each other, in the proportion of the squares of their corresponding or homologous sides; that is, (fig. 34) the area of the triangle ABC, will be to that of the triangle DEF, as the square of AB is to the square of DE, or as the square of BC to the square of EF.

PROP. XXII. fig. 2, Plate 2. To bisect a To bisect a given right line, AB, that is, to cut it into two equal parts. Upon

the extremity A, as a centre, with a radius or opening of the compasses greater than half of AB, make arches on each side of that line, as a b, cd; from B, as a centre, with the same radius, describe the arches ef, gh, intersecting the former arches in the points I and K: then lay a ruler, or draw a line from I to K, cutting AB in the point O; so will AB be bisected at O, that is, AO will be equal to OB. For the points I and K being found by intersecting arches described with the same radius, these points must be equally distant from the extremities A and B ; and

a right

a right line joining I and K, will also, in every other point, be equally distant from A and B; consequently the point O, where IK cuts AB, must be equally distant from A and B, that is, the line AB must be cut into two equal parts at the point O.

PROP. XXIII. fig. 3. From a given point, C, in the right line, AB, to draw a right line, CF, which shall be perpendicular to AB. Take the points D and E in AB, equally distant from the point C; then from D and E, with any radius greater than DC or CE, make the intersection F, and join FC: then will FC be perpendicular to AB, and it is drawn at the given point C, which was the thing required to be done. See Prop. 22.

PROP. XXIV. fig. 4. From a given point, C, to let fall a perpendicular on a given line, AB. From C, as a centre, with a radius reaching beyond AB, describe an arch cutting AB in two points; and from these points, with any convenient radius, make the intersection D on the side opposite to C: a line drawn from C towards D, till it touch AB, in the point E, will be perpendicular to AB; and it is drawn from the point C, as was required

to be done.

PROP. XXV. fig. 5. To erect a perpendicular on the extremity B, of a given right line, AB. Choose any point, C, such that a perpendicular let fail from it, would come within the point B: from C, with the radius CB, describe the arch DBE, cutting AB in D; draw a line joining D and C, and produced till it cut the arch in E; then the segment, or arch, DBE, being a semicircle, the angle at B is a right angle, consequently the line EB is drawn perpendicular to the line AB, at the point B, as was required.

PROP.

PROP. XXVI. fig. 6. At a given point, D, of a given right line, DE, to make an angle which shall be equal to a given angle, CAB. On A, as a centre, with any radius. Ab, draw the arch ba; on D, as a centre with the same radius, draw an arch beginning at e: next take in the compasses the opening of the arch ba, and set the same. up from e tod; and drawing the line D d F, this line will form with the given line DE, the angle at D equal to the given angle at A.

PROP. XXVII. fig. 7. Through a given point, C, to draw a line parallel to a given right line, AB. From any convenient point in AB, as B, with the radius BC, describe the arch CD; from C, with the same radius describe the arch BE: take the distance CD, and set it up from B to E; and draw the line CE, which will be parallel to the given line AB: for the line CB falling upon the two lines CE and AB, in such a way that the alternate angles, ECB and CBD, are equal, (Prop. 26,) it follows from what was formerly said of the proportion of parallel lines, that CE must be parallel to AB, and it is drawn through the point C, as was required.

PROP. XXVIII. fig. 8. To bisect a given angle, ABC. On B, as a centre, describe the arch AC: from these points make the intersection D: then join DB, and this line will bisect the given angle ABC.

The same line, DB, will also bisect the arch AC.

PROP. XXIX. fig. 9. To construct a triangle whose sides shall be equal to three given right lines, as AB, CD, EF. Draw the line GH, and make it equal to AB: with CD, as a radius, describe an arch at L: with EF, as a radius, describe another arch, cutting the former in the

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t

point L, and join LG, and LH; then will the triangle GLH have its sides respectively equal to the three given right lines:

PROP. XXX. fig. 10. To construct, a parallelogram, whose opposite sides shall be equal to two given right lines, AB and CD, and which shall contain an angle equal to the given angle at E. Draw FG, making it equal to AB: at F make the angle HFG, equal to the angle at E, (Prop. 26); make FH equal to CD; on H, with a radius equal to FG or AB, and on G, with a radius equal to FH or CD, make an intersection at I, and draw HI, GI; then will the parallelogram, FHIG, have the opposite sides HI and FG, each equal to AB; FH and GI, each equal to CD; the angles HFG and HIG, each equal to the given angle at E.

and

PROP. XXXI. fig. 11 and 12. Through a given point to draw a tangent to a given circle. This problem admits of two solutions: the one, when the given point A is in the circumference, as in fig. 11; and the other, when the point A is without the circle, as in fig. 12. In the first case, draw AC to the centre, and on that radius, at A, erect a perpendicular AB, which being produced to D, will touch the circle at the point A, but afterwards fall without it, and will consequently be a tangent drawn through the given point.

But in the second case, where the given point A is without the circle, draw AC to the centre, and bisecting this line in the point B, describe the circle DCE, or so much of it as may cut the given circle in the points D and E: draw the lines AD and AE, and these will both be tangents to the given circle: for if the radii, CD and CE, be drawn, the two angles ADC, AEC, being angles in a semicircle, are right angles; consequently AD and AE being at right angles to the radii, will never fall within the circle, but

touch