PROPOSITION XX. THEOREM. (Converse of Prop. IX.)

In any triangle the longer side is opposite to the greater angle.

In the triangle ABC let the angle C be greater than the angle B, then will the side AB be longer than the side AC.

B

For, if AB were equal to AC, the angle C would be equal to the angle B. If ÅB were shorter than AC, then would the angle C be less than the angle B (Prop. XIX. Cor.). As, therefore, AB can be neither equal to, nor shorter than AC, it must necessarily be longer. The longer side, therefore, is opposite to the greater angle. Cor. I. Therefore the shorter side is opposite to the less angle.

Cor. 2. In the right angled triangle the hypothenuse is the longest side (Prop. XVI. Cor. 4.).

PROPOSITION XXI. THEOREM.

Any two sides of a triangle are together longer than the third side.

The two sides AB, AC, for instance, of the triangle ABC are together longer than the third side BC. For, let BA be produced till AD be equal to AC, and let DC be joined.

Then the angle ACD being equal to the angle D (Prop. IX.), the angle BCD must be greater than the angle D; consequently the side BD is longer than the side BC (Prop. XX.). But BD is equal to BA and AC together; therefore the two sides AB, AC, are together longer than BC.

Cor. Therefore AC is longer than the difference between. AB, BC; and AB longer than the difference between AC, BC; that is, any side of a triangle exceeds the difference between the other two.

The perpendicular drawn from a point to a straight line is shorter than any other line drawn thereto from the same point; and those lines which meet

the proposed line at equal distances from the perpendicular are themselves equal, and the more remote from the perpendicular the point of meeting is, the, longer is the line drawn.

Let PF be perpendicular to the straight line AB, and let PD, PE, intercept equal distances DF, EF; let also any other oblique line PC be drawn.

C D F E

B

First, since the angle PFE is right, PE is longer than PF (Prop. XX. Cor. 2.); therefore the perpendicular is shorter than any oblique line.

Again, since DF, FP, and the included angle are equal to EF, FP, and the included angle PD is equal to PE (Prop. VIII.); hence lines drawn from Pintercepting equal distances from the perpendicular are equal.

Lastly, because the triangle PDE is isosceles, as we have just shown, the exterior angle PDC is greater than the inward adjacent angle PDE (Prop. XVI. Cor. 7.), and this last is greater than the angle PCD (Prop. XVI.); therefore PDC being greater than PCD, PC is longer than PD (Prop. XX.); that is, the more remote from the perpendicular any oblique line falls, the longer it is.

Cor. 1. It follows from this last case that two equal straight lines cannot be drawn from a point to a line, and fall upon the same side of the perpendicular from that point to the line; and that, therefore, it is impossible to draw three equal straight lines from the same point to a given straight line.

Scholium.

The converse of this proposition immediately follows, that is, First, The shortest line that can be drawn from a point to a straight line is a perpendicular thereto; for by the first part of the preceding demonstration, if this were not the perpendicular, it could not be the shortest line.

Secondly, If equal lines be drawn from a point to a line, the distances intercepted between them and the perpendicular from that point will be equal; for, by the third case of the above, if the distances intercepted were unequal, the lines drawn would be also unequal.

And, lastly, If unequal lines be drawn, the longer shall fall more remotely from the perpendicular; for, if it were less remote it would be shorter, if equally remote, equal, as we have already proved.

Cor. 2. Hence, if from a point to a line two lines be drawn, then, that which is not shorter than the other shall exceed any line drawn between them.

Cor. 3. And the shorter of two lines so drawn shall be shorter than any line drawn between them.

Cor. 4. A line drawn from the middle of another to a point equally distant from its extremities, is a perpendicular thereto; for a perpendicular, from the point to the line, is equally distant from its extremities (Schol.).

Cor. 5. Therefore, if through two points, of which each is equally distant from the extremities of a straight line, a second line be drawn, it shall be perpendicular to the first; for, by last corollary, a line from the middle of the former to either point is a perpendicular.

Cor. 6. It, moreover, follows that two right angled triangles are equal, when the hypothenuse and one side in the one triangle are respectively equal to the hypothenuse and a side in the other.

PROPOSITION XXIII. THEOREM.

If two sides of one triangle be respectively equal to two sides of another, but include a greater angle; the third side of the former shall exceed the third side of the latter.

Let ABC, DEF, be two triangles having any two sides, as AB, AC, in the one, respectively equal to two sides DE, DF, in B the other, while the angle included by the former is greater

D

CE

H

than that included by the latter; then will the third side BC, of the former, be longer than the third side EF, of the latter. Of the two sides AB, AC, let AC be that which is not shorter than the other, let the line AGH make an angle with AB equal to the angle D, let AH be equal to DF or AC, and let BH, HC, be drawn.

Since AC is not shorter than AB, it is longer than AG (Prop. XXII. Cor. 2.); therefore, as AH is equal to AC, the extremity H must fall below the line BC. The angles ACH, AHC, are equal (Prop. IX.); hence the angle BCH is less than the angle AHC, and, therefore, necessarily less than BHC; hence the side BC is longer than the side BH (Prop. XX.); but BH is equal to EF, because the sides AB, AH, and the included angle are respectively equal to DE, DF, and the included angle (Prop. VIII.); consequently BC is longer than EF.

PROPOSITION XXIV.

THEOREM. (Converse of Prop. XXIII.)

If two sides of one triangle be respectively equal to two sides of another, while the third side of the former is longer than that of the latter, the angle included by the former two sides shall exceed that included by the latter two.

In the triangles ABC, DEF, let the two sides AB, AC, be equal respectively to the two sides DE, DF; while the side BC ex- B ceeds the side EF, the angle A will exceed the angle D.

E

For the angle A cannot be equal to the angle D, for then the side BC would be equal to the side EF (Prop. VIII.); nor can it be less, for then BC would be less than EF (Prop. XXIII.). As, therefore, the angle A can be neither equal to, nor less than, the angle D, it must necessarily be greater.

Two triangles are equal which have the three sides of the one respectively equal to the three sides of the other.

For the angle included between any two sides in the one triangle must be equal to the angle included by the two corresponding sides in the other; since, if it were unequal, the opposite sides would be unequal (Prop, XXIII.), which is contrary to the hypothesis; therefore the three angles in the one triangle are respectively equal to the three angles in the other.

Cor. From this, and Prop. VIII, it follows that one quadrilateral is equal to another, if the sides of the one are respectively equal to the sides of the other: and the angle included by any two sides of the one also equal to the angle contained by the two corresponding sides of the other.

PROPOSITION XXVI. THEOREM.

Two triangles are equal, if two sides, and an opposite angle in one are respectively equal to two sides; and a corresponding opposite angle in the other, provided the other opposite angles in each triangle are either both acute, or both obtuse.

In the triangles ABC, DEF, let the sides AB, AC, be respectively equal to the sides DE, DF, and let the angle C, opposite to the side. A B, be equal to the angle F, opposite to the side DE; then will BC be equal to EF, provided the angles B and E are either both acute, or both obtuse.

First, let B and E be acute, then, if the equality of BC and EF be denied, one of them as

EF must be longer than

B

CE

D

the other. Let, then, FG be taken equal to BC, and draw DG, which will be equal to AB (Prop. VIII.), and, therefore, equal to DE; consequently the angle E is equal to the angle DGE (Prop. IX.); DGE is, therefore, an acute angle, but this angle, together with DGF, make up two right angles (Prop. III.); DGF is, therefore, an obtuse angle. But, since the triangles ABC, DGF, are equal, the angle DGF must be equal to the angle B, and, therefore, acute, which is impossible; so that FG cannot be equal to BC, and the demonstration would have been the same had BC been supposed longer than EF; these two sides are, therefore, equal.

Next, let the angles B and

E be obtuse; then, if EF be supposed longer than BC, produce the latter beyond the G vertex B, till CG be equal to

E

EF:-join AG. Then, as before shown, AG is equal to DE, or to AB, and the angle AGC, which is equal to the angle ABG (Prop. IX.), is acute, since ABC is obtuse; but the same angle must be obtuse, because the triangles AGC, DEF, are equal (Prop. VIII.) which is impossible; whence EF cannot be longer than BC, and had BC been supposed longer than EF a similar absurdity would obviously have followed; hence in this case also the sides BC, EF, are equal, and, therefore, (Prop. XXV.) the triangle ABC is equal to the triangle DEF.

The opposite sides and angles of a rhomboid are equal. Let ABCD be a rhomboid, the opposite sides and angles are equal.

Draw the diagonal AC, then, since AB, DC are parallel, the alternate angles BAC, DCA, are equal (Prop. XIV.), and because AD, BC, are also parallel, the alternate angles

A