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PROPOSITION VIII.

THEOREM. When two triangles have the three sides of the one respectively equal to the three sides of the other, the triangles will be identical, and equal in all respects.

Let the two triangles ABC ABD have their sides respectively equal, namely, AB equal to AB, AC equal to AD, and BC equal to BD; then will these triangles be identical.

D

For, conceive the two triangles to be joined together by their longest equal sides, and draw the line CD; then in the triangle ACD, since AC is equal to AD, we have the angle ACD equal to the angle ADC. (Prop. v.) In like manner, in the triangle BCD, since BC is equal to BD, we have the angle BCD equal to the angle BDC. Hence the angle ACB, which is the sum of ACD and BCD, is equal to the angle ADB, which is the sum of ADC and BDC. Since, then, in the triangle ACB, we have the two sides AC and BC, and their included angle ACB, equal respectively to the two sides AD and BD, and their included angle ADB, of the triangle ADB, it therefore follows that these triangles are identical. (Prop. III.)

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PROBLEM.

PROPOSITION IX.

Three straight lines, A, B, C, each of which is less than the sum of the two others, being given, to construct a triangle whose sides shall be respectively equal to them.

Make the line DF equal to the line A; and with D and F as centres, and with radii equal respectively to the lines B and C, describe arcs intersecting at the point G. (Post. III.) Join DG, FG, (Post. I,) and the triangle DGF will be the triangle

A B C

D

required, since the three sides are equal to the three lines A, B, C.

Scholium. It is obvious that the arcs described from D and F as centres, will intersect in two points G and H, thus giving two triangles, DGF and DHF; but these two triangles, having the three sides of the one equal to the three sides of the other, are identical. (Prop. VIII.)

If two of the given lines are equal, the triangle will be isosceles; when all the lines are equal, it will be equilateral.

(35.) THEOREM. If, on the three sides of any triangle, equilateral triangles be constructed externally to the given triangle, then will the straight lines drawn from the vertices of the equilateral triangles to the opposite angles of the given triangle be equal.

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.

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. 11,) 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.

(36.) 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.

For, on DH construct the equilateral triangle DHK, and join GK; then comparing the two triangles

ABC

D

DFH and 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. 111,) and consequently GK is equal to DF. Hence the three lines GK, 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

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ing 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. VIII,) the angle FDG will be equal to the angle BAC.

PROBLEM.

PROPOSITION XI.

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 tri

A

B

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

PROPOSITION XII.

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

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 triangle 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.

H

B

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