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- в 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.

PROPOSITION XXV. THEOREM.

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.

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 AB, 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

B

CE

G

D

EF must be longer than 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 & vertex B, till CG be equal to

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

D

A

B

DAC, BCA, are likewise equal; hence the two angles BAC, DAC, are together equal to the two angles DCA, BCA; that is, the opposite angles BAD, DCB, are equal. Again, since the angles BAC, BCA, and the interjacent side of the triangle ABC are respectively equal to the angles DCA, DAC, and the interjacent side of the triangle CDA, the triangles are equal (Prop. XI.); therefore the side AB is equal to the side CD, the side BC to DA, and the angle B to the angle D; hence, in a rhomboid, the opposite sides and angles are equal. Cor. 1. From this proposition, and Cor. 3. to Prop. XVII., it follows, that if one angle of a rhomboid be right, all the angles will be right.

Cor. 2. Therefore in the rectangle and square (see Definitions) all the angles are right, and in the latter all the sides are equal.

Cor. 3. The diagonal divides a rhomboid into two equivalent triangles.

Cor. 4. Parallels included between two other parallels are equal.

PROPOSITION XXVIII.

THEOREM. (Converse of Prop.
XXVII.)

If the opposite sides of a quadrilateral be equal, or if the opposite angles be equal, the figure will be a rhomboid.

In the quadrilateral ABCD let the opposite sides be equal, the figure will be a rhomboid.

A

D

B

Let the diagonal AC be drawn, then the triangles ABC, ADC, are equal, since the three sides of the one are respectively equal to those of the other (Prop. XXV.); therefore the angles BAC, DCA, opposite the equal sides BC, DA, are equal; therefore DC is parallel to AB; the angles ACB, CAD, opposite the equal sides BC, DA, are also equal; BC is, therefore, parallel to AD (Prop. XII.); hence ABCD is a rhom

boid.

Next, let the opposite angles be equal.

Then the sum of the angles BAD, ADC, must be equal to the sum of the angles DCB, CBA; therefore each sum is equal to two right angles (Prop. XVII. Cor. 3.); therefore AB, DC, are parallel (Prop. XII. Cor. 3.). For similar reasons AB, BC, are parallel; therefore the figure is a rhom

boid.

If two of the opposite sides of a quadrilateral are both equal and parallel, the figure is a rhomboid.

In the quadrilateral ABCD (preceding diagram), let AB be equal and parallel to DC, then will ABCD be a rhomboid.

For the diagonal AC makes the alternate angles BAC, DCA, equal (Prop. XIV.); so that in the triangles ABC, CDA, two sides, and the included angle in each are respectively equal; these triangles are, therefore, equal (Prop. VIII.), the angle ACB is, therefore, equal to the angle CAD; hence AD is parallel to BC (Prop. XII.), and the other two sides are parallel by hypothesis; therefore ABCD is a rhomboid.

Cor. If, in addition, the parallel sides be each equal to a third side, the rhomboid will be either a rhombus or a square, according as it has, or has not, a right angle.

Scholium.

It has been proved (Prop. VIII.) that two triangles are equal when two sides, and the included angle in the one are respectively equal to two sides, and the included angle in the other; we may now infer further, that two triangles are equivalent, or equal in surface, when two sides of the one are respectively equal to two sides of the other, and the sum of the included angles equal to two right angles.

For, let the triangles ADC, BCD, having two sides AD, DC, in the one equal to the two BC, CD, in the other be placed as in the margin, a side of the one coinciding with the equal side in the other; let also the included angles ADC, BCD, be together equal to two right angles, and let AB, BD, be drawn.

A

B

Then, since the angles ADC, BCD, are together equal to two right angles, the lines AD, BC, are parallel (Prop. XII. Cor. 3.), but they are also equal by hypothesis; hence, by the above proposition, the figure ABCD is a rhomboid; now, the triangle ADC is half the rhomboid (Prop. XXVII. Cor. 3.), so also is the triangle BCD; these triangles are, therefore, equivalent.