THEOREM 18.

The difference of any two sides of a triangle is less than the third side.

Let ABC be the A, and let AC be greater than AB; then we are to prove that AC-AB is less than BC.

As a straight line is the shortest distance between two points,

Therefore,

AB+BC > AC.

From these unequals subtract the equals AB=AB, and we have BC AC-AB. (ax. 5). Q. E. D.

When two triangles have all three of the sides in one triangle equal to all three in the other, each to each, the two triangles will be identical, and have equal angles opposite equal sides.

In two triangles, as ABC and ABD, on the supposition that the side AB of the one=AB of the other, AC AD, and BC BD, we are to demonstrate that the angle ACB the angle ADB, BAC= BAD, and ABC=ABD.

Conceive the two triangles to be joined together by their longest equal sides, and draw the line CD.

Then, in the triangle ACD, because the side AC is equal to AD by (hy.), the angle ACD is equal to the angle ADC (th. 15). In like manner, in the triangle BCD, the angle BCD is equal to the angle BDC, because the side BC is equal to BD. Hence, then, the angle ACD being equal to the angle ADC, and the angle BCD to the angle BDC, by equal additions the sum of the two angles ACD, BCD, is equal to the sum of the two ADC, BDC (ax. 2); that is, the whole angle ACB is equal to the whole angle BDA.

Since then the two sides, AC, CB, are equal to the two sides AD, DB, each to each, by (hy.), and their contained angles ACB, ADB, also equal, the two triangles ABC, ABD, are identical (th. 13), and have their other angles equal, the angle BAC to the angle BAD, and the angle ABC to the angle ABD. Q. E. D.

If there be two triangles which have the two sides of the one equal to the two sides of the other, each to each, and the included angles unequal, the third sides will be unequal, and the greater side will belong to the triangle which has the greater included angle.

Let ABC be one A, and ACD the other A. Let AB and AC of the one be equal to AD and

AC of the other A. But the angle BAC greater than the angle DAC; then we are to prove that the base BC is greater than the base CD.

A perpendicular is the shortest line that can be drawn from any point to a straight line; and if other lines be drawn from the same point to

the same the same straight line, the greater will be at a greater distance from the perpendicular; and lines at equal distances from the perpendicular. on opposite sides, are equal.

Let A be any point without the line DE; and let AB be the perpendicular; AC, AD, and AE oblique lines: then, if BC is less than BD, and BC=BE, we are to show,

1st. That at AB is less than AC. 2d. AC less than AD. 3d. AC-AE.

In the triangle ABC, as AB is perpendicular by (hy.), the angle ABC is a right angle; then, as it requires the other two angles of the triangle (th. 11) to make another right angle, the angle ACB, is less than a right angle; and as the greater side is always opposite the greater angle, AB is less than AC; and as AC is any line differing from AB, therefore AB is the least of any line drawn re AB is the least of any line drawn from A.

2d. As the two angles ACB and ACD (th. 1) make two right angles, and ACB less than a right angle, therefore ACD is greater than a right angle; consequently, the D is less than a right angle; and, therefore, in the AACD, AD is greater than AC, or AC is less than AD.

3d. In the As ABC and ABE, AB is common, and CB=BE, and the angles at B, right angles; therefore, by (th. 15) AC=AE. Q. E. D.

THEOREM 21.

The opposite sides, and the opposite angles of any parallelogram, are equal to each other.

Let ABDC be a parallelogram. Then we

are to show that AB CD, AC=BD, the an

gle A=D, and the angle ACD=ABD.

Draw a diagonal, as CB; then, because

AB and CD are parallel, the alternate an- A

gles ABC and BCD (th. 5) are equal. For the same reason, as AC and BD are parallel, the angles ACB and CBD are equal. Now, in the two triangles ABC and BCD, the side CB is common,

and

The JACB=_ CBD
and BCD=JABC

(1)
(2)

Therefore, the third angle A= the third angle D (th. 11), and by (th. 13) the two As are equal in all respects; that is, the sides opposite the equal angles are equal; or, AB= CD, and AC-BD. By adding equations (1) and (2), (ax. 2), we have the angle ACD = the angle ABD; therefore, the opposite sides, &c. Q. E. D.

Cor. 1. As the sum of all the angles of the quadrilateral is equal to four right angles, and the angle A is always = to the opposite angle D; if, therefore, A is a right angle, D is also a right angle, and all the angles are right angles.

Cor. 2. As the angle ABD, added to the angle A, gives the same sum as the angles of the AACB; therefore, the two adjacent angles of a parallelogram make two right angles; and this corresponds with the 4th point of theorem 12.

THEOREM 22.

If the opposite sides of a quadrilateral are equal, they are also parallel, and the figure is a parallelogram.

Let ABDC represent any quadrilateral,

and on the supposition that AC=BD, and AB CD, we are to prove that AC is parallel to BD, and AB parallel to CD.

Draw the diagonal CB; then we have two A triangles ABC, and CDB, which have the common side CB; and AC of the one=BD of the other, and AB of the one- CD of the other; therefore by (th. 19) the two As are equal, and the angles equal, to which the equal sides are opposite; that is, the angle A CB the angle CBD, and these are alternate angles; and, therefore, by (th. 5), AC is parallel to BD; and because the angle ABC= BCD, AB is parallel to CD, and the figure is a parallelogram. Q. E. D.

=

Cor. In this, and also in (th. 21), we proved that the two As which make up the parallelogram are equal; and the same would be true if we drew the diagonal from A to D; and in general we may say, that the diagonal of any parallelogram bisects the parallelogram.

THEOREM 23.

The lines which join the corresponding extremities of two equal and parallel straight lines, are themselves equal and parallel; and the figure thus formed is a parallelogram.

On the supposition that AB is equal and parallel to CD (see last figure), we are to show that AC will be equal and parallel to BD; and that will make the figure a parallelogram.

Join CB; then because AB and CD are parallel, and CB joins them, the alternate angles ABC and BCD are equal, and the side AB CD, and CB common to the two As ABC and CDB; therefore by (th. 13) the two triangles are equal; that is, AC= BD, the angle A=D, and ACB=CBD; hence, AC is also parallel to BD; and the figure is a parallelogram. Q. E. D.

THEOREM 24.

Parallelograms on the same base, and between the same parallels,

are equal in surface.

Let ABEC and ABFD be two parallelograms on the same base AB, and between the same parallel lines AB and CD; then we are to show that these two parallelograms are equal.

Now CE and FD are equal, because they are each equal to AB (th. 21); and if from the whole line CD we take, in succession, CE and FD, there will remain (ax. 3) ED=CF; but EB CA, and AF=BD (th. 21); hence we have two As, CAF and EBD, which have the three sides of the one equal to the three corresponding sides of the other, each to each; and therefore by (th. 19) the two As CAF and EBD are equal. If from the whole figure we take away the ▲ CAF, the parallelogram ABDF remains; and if from the whole figure the other triangle EBD be taken away, the parallelogram ABEC will remain; that is, from the same quantity, if equals are taken (ax. 3), equals will be left; or the parallelogram ABDF-ABEC. Q. E. D.