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ulars be drawn, these perpendiculars will be parallel, for each will be perpendicular to both parallels (Prop. 9, Cor. 2).

PROP. XI. THEOREM.

If two straight lines be parallel to a third, they will be parallel to each other.

Let CD, EF, be parallel to AB; then we have to prove that they will be parallel to each other.

A

C

E

Fig. 13.

H

B

D

·F

Draw IH perpendicular to AB, it will also (by Prop. 9, Cor. 2) be perpendicular to both CD and EF; hence these lines, being both perpendicular to the same line, are parallel to each other (Prop. 10, Cor. 2).

PROP. XII. THEOREM.

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If one side of a triangle be produced, the exterior angle, so formed, will be equal to the sum of the two in

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formed, will be equal to the sum of the interior and opposite angles BAC, BCA.

From the point B draw the line BE parallel to AC; then will the angles EBD, BAC, be equal (Prop. 9, Cor. 1), and the angles CBE, BCA, be equal (Prop.

9); hence CBD, which is the sum of the two angles CBE, EBD, is equal to the sum of the two angles ВАС, ВСА.

Cor. 1. The sum of the three angles of a triangle is equal to two right angles; for CBD+CBA is equal to two right angles (Prop. 1); now we have just seen that CBD is equal to BAC+BCA; hence BAC+BCA+ CBA is equal to two right angles.

Cor. 2. Hence, if two angles in one triangle be equal to two angles in another triangle, the third angle in the one will be equal to the third angle in the other.

Cor. 3. In any triangle there can be but one right angle; for if there could be two, the third angle would be nothing.

Cor. 4. In a right-angled triangle the sum of the two acute angles is equal to one right angle.

PROP. XIII. THEOREM.

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In any polygon, the sum of all the angles is equal to as many times two right angles as the figure has sides, less four right angles.

Fig. 15.

From any point O within the polygon, draw lines to the vertices of all the angles. There will evidently be formed as many triangles as the figure has sides, and the sum of all the angles of the polygon will be less than the sum of all the angles of the triangles, by the angles formed at the point O. Now

the sum of all the angles of each triangle is equal to two right angles (Prop. 12); and the sum of all the angles

formed about O is four right angles (Prop. 1, Cor. 3); hence the sum of all the angles of the polygon is equal to as many times two right angles as the figure has sides, less four right angles.

Cor. For example, the sum of all the angles of a sixsided figure is eight right angles; since it is equal to as many times two right angles as the figure has sides, (or twelve right angles,) less four. If all the six angles are equal to each other, each of them will be the sixth part of eight right angles, or = of a right angle.

PROP. XIV. THEOREM.

In a triangle, the angle which is opposite the longest side is the largest.

In the triangle ABC, let the side AC be longer than

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AD, of the triangle ABD, are equal, the angles ABD, ADB, must be equal (Prop. 6); but the angle ADB, being an exterior angle of the triangle DBC, is equal to the sum of the angles DBC, DCB, (Prop. 12), and is therefore larger than C; hence ABD must be larger than C; and ABC, which exceeds ABD, must be still larger than C. In a similar manner, if the side AC is taken longer than BC, may we prove the angle ABC to be larger than A.

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SUM OF TWO SIDES OF TRIANGLES. [BOOK I.

PROP. XV. THEOREM.

(Converse of Prop. XIV.)

In a triangle, the side which is opposite the largest an

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If AC is not longer than BC, it is either equal to it, or less than it; if it were equal to it, the angle B would be equal to the angle A (Prop. 6); and if it were less than it, the angle A would be larger than B (Prop. 14) ; both of these results are contrary to the supposition; hence AC must be longer than BC.

Cor. Hence in a right-angled triangle the hypotenuse is the longest side. (Prop. 12, Cor. 3.)

PROP. XVI. THEOREM.

The sum of any two sides of a triangle is longer than the third side.

In the triangle ABC, we have to prove, for example, that AB+BC is longer than AC.

Produce AB until

BD is equal to BC, A

Fig. 18.

and draw CD; then the angle BCD-D (Prop. 6); therefore ACD, which is larger than BCD, is also larger than D; hence AD is longer than AC (Prop. 15); but

BOOK I.] PERPENL.CULARS AND OBLIQUE LINES, 23

AD is equal to AB+BC; hence the sum of the two sides AB, BC, is longer than the third side AC; the same thing may be proved of the sum of any other two sides.

PROP. XVII. THEOREM.

A perpendicular is the shortest distance from a point to a straight line.

Let A be the point, BC the straight line; we have to prove that the perpendicular

AD is shorter than any other line which can be drawn from A to BC.

Draw, for example, AE; then is AD<AE (by Prop. 15, Cor.)

Fig. 19.

B

E

'D

PROP. XVIII. THEOREM.

If a perpendicular be drawn to a given straight line, and, from any point on the perpendicular, oblique lines be drawn to the given line, so as to cut off equal distances from the foot of the perpendicular, these oblique lines will be equal.

Let AB be the given straight line, CD the perpendicular, and DE=DF;

then we have to prove that the oblique lines CE, CF, drawn from any point C of the perpendicular, are equal.

C Fig. 20.

B

E

D

F

In the two right-angled triangles EDC, FDC, we have the sides ED, DC, and the included angle EDC, of the one, respectively

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