Let ABC be the given triangle, upon whose sides the equilateral triangles ADB, BFC, CGA, are described; then will the lines DC, FA, GB, be equal.

A

B

Comparing the two triangles ABF, DBC, we have AB = DB; BF = BC, being sides of equilateral triangles. The angles CBF, ABD are equal, being angles of equilateral triangles; to each of these angles add the angle ABC, and we have the angle ABF DBC. Therefore the two triangles ABF and DBC have the two sides and the included angle of the one equal to the two sides and the included angle of the other; they are therefore identical (Prop. II), and consequently DC is equal to FA.

=

By comparing the triangles ACF and GCB, it may be shown that they also are equal, and consequently FA is equal to GB.

It might, moreover, be shown that these lines all intersect in the same point.

(28.) PROBLEM. Given the lengths of three lines drawn from a point within an equilateral triangle, to the three corners; to find the side of the triangle.

Let A, B, C, be the three given lines. With these lines construct the triangle DFG (Prop. 1x); upon either of the sides, as FG, construct the equilateral triangle FGH; join DH, and it will be a side of the equilateral triangle sought.

А В С

D

For, on DH construct the equilateral triangle DHK, and join GK; then comparing the two triangles DFH & KGH, we see that DH and FH of the one are respectively equal to KH and GH of the other, being sides of equilateral triangles; the angle FHG is equal to DHK,each being an angle of an equilateral triangle (Prop. v, Cor. 2); from each take the angle DHG, and we have FHD equal to GHK; therefore the two triangles DFH and KGH are equal (Prop. II), and consequently GK is equal to DF. Hence the three lines GK,

C

GD, GH, are respectively equal to the three DF, DG, GF, which are equal to the given lines A, B, C; which proves DH to be the side of the equilateral triangle sought.

PROPOSITION X.

PROBLEM. At a given point D, in a given line DF, to make an angle equal to a given angle BAC.

With any radius describe arcs from

A

B

H

F

A & D as centres (Post. III); the first BC, meeting AB, AC, at B and C; and the second FH meeting DF at F. With F as a centre, and a radius equal to the distance from B to C, describe an arc (Post. III) to meet FH at G. The line DG being drawn, will make the angle FDG equal to BAC.

This is an application of Prop. IX, and the equality of the angles will follow from Prop. VIII. For, drawing lines from B to C, and from F to G (Post. I), we have the three sides of the triangle ABC equal to the three sides of the triangle DFG; therefore (Prop. vii) the angle FDG will be equal to the angle BAC.

PROPOSITION XI.

B

PROBLEM. To bisect a given angle ABC. Take any equal distances BA, BC, on the sides containing the angle ; and with A and C as centres, and any equal radii, describe arcs intersecting each other at D; then, BD being drawn, it will bisect the angle ABC. For, drawing AD, CD, the three sides of the triangle ABD are

equal respectively to the three sides of the triangle CBD; hence (Prop. vii) the angles ABD and CBD are equal.

PROPOSITION XII.

PROBLEM. Through a given point C, in a given line AB, to draw a perpendicular.

1

Take equal distances CD, CF, on each side of the given point; and with any equal radii, describe arcs (Post. III) meeting at G. Join GC (Post. I), and it will be the perpendicular required.

D

C F

For, conceive DG, FG, to be drawn; then the equality of the sides of the triangles DGC, FGC, give the angles at C equal (Prop. viii), and hence GC is perpendicular to AB (Def. X).

PROPOSITION XIII.

PROBLEM. To bisect a given straight line AB.

With A and B as centres, with any convenient equal radii, describe arcs (Post. III) intersecting at C and D. Draw CD (Post. I), and it will be perpendicular to AB, and will bisect it at the point F.

A

B

For, by joining AC and BC, AD and BD, we shall have two triangles CAD and CBD, with all the sides of the one equal respectively to all the sides of the other; consequently the angle ACF is equal to BCF (Prop. ví). Hence since the line CF bisects the vertical angle of the isosceles triangle ACB, it bisects the base AB (Prop. v, Cor. 1).

PROPOSITION XIV.

PROBLEM. From a given point A without a given line BC, to draw a line perpendicular to BC.

With A as a centre, with any convenient radius, describe an arc (Post. III) cutting BC in the two points D and F; and with D and F as centres, and with equal radii, describe arcs (Post. III) intersecting at G. Then AG being drawn, cut

B

A

ting BC at H, will be the perpendicular required.

For, joining DA and FA, DG and FG, we have all the sides of the triangle AÐG equal respectively to all the sides of the triangle AFG; consequently the angle DAH is equal to to FAH (Prop. vín); and since AH bisects the vertical angle of the isosceles triangle DAF, it is perpendicular to the base DF (Prop. v, Cor. 1).

(29.) Let ABC be a triangular field, and S a well within it; it is required to divide the field into two parts by a line passing through S, so that the distances A AF, AG shall be equal.

G

K

L

B

F

Solution. Draw the line AK, bisecting the angle BAC (Prop. x1); and through the point S draw GF perpendicular to AK (Prop. xiv), and it will be the line required.

For, comparing the two triangles ALF, ALG, we have the angle LAF equal to the angle LAG by construction; the angle ALF equal to the angle ALG, each being a right angle; and the side AL common hence we have two angles and the interjacent side of the one triangle equal to two angles and the interjacent side of the other; consequently the triangles are equal, and AF = AG (Prop. 1v).

PROPOSITION XV.

THEOREM. The greater side of every triangle is opposite the greater angle; and the greater angle is opposite the greater side.

In the triangle ABC, let the side AB be greater than the side AC; then will the angle ACB opposite the greater side AB be greater than the angle ABC opposite the less side.

A

C

B

D