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For, by joining AC and BC, AD and BD, we shall have two triangles, CAD and CBD, with all the sides of the one equal respectively to all the sides of the other; consequently the angle ACF is equal to BCF (Prop. VIII.) Hence, since the line CF bisects the vertical angle of the isosceles triangle ACB, it bisects the base AB, (Prop. v, Cor. 1.)

PROPOSITION XIV.

PROBLEM. From a given point A without a given line BC, to draw a line perpendicular to BC.

B

A

With A as a centre, with any convenient radius, describe an arc (Post. III,) cutting BC in the two points D and F; and with D and F as centres, and with equal radii, describe arcs (Post. III,) intersecting at G. Then AG being drawn, cutting BC at H, will be the perpendicular required.

For, joining DA and FA, DG and FG, we have all the sides of the triangle ADG equal respectively to all the sides of the triangle AFG; consequently the angle DAH is equal to FAH, (Prop. VIII;) and since AH bisects the vertical angle of the isosceles triangle DAF, it is perpendicular to the base DF (Prop. v, Cor. 1.)

(37) Let ABC be a triangular field, and S a well within it; it is required to divide the field into two parts by a line passing through S, so that the distances AF, AG shall be equal.

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K

L

F

Solution. Draw the line AK, bisecting the angle BAC, (Prop. xi;) and through the point S draw GF perpendicular to AK, (Prop. xiv,) and it will be the line required.

For, comparing the two triangles ALF, ALG, we have the angle LAF equal to the angle LAG by construction; the angle ALF equal to the angle ALG, each being a right-angle; and the side AL common: hence we have two angles and the interjacent side of the one triangle equal to two angles and the interjacent side of the other; consequently the triangles are equal, and AF=AG, (Prop. iv.)

PROPOSITION XV.

THEOREM. The greater side of every triangle is opposite the greater angle; and the greater angle is opposite the greater side.

In the triangle ABC, let the side AB be greater than the side AC; then will the angle ACB opposite the greater side AB be greater than the angle ABC opposite the less side.

A

For, from the greater side AB take a portion AD equal to the less side AC, and join CD. Now, since the angle ADC is an exterior angle in reference to the triangle CDB, it is greater than the interior angle ABC, (Prop. vi;) but since the side AD is equal to the side AC, the angle ACD is equal to ADC, (Prop. v,) and consequently the angle ACD is greater than ABC; and since the angle ACD is only a part of the angle ACB, much more, then, must the whole angle ACB be greater than ABC. Again, conversely, if the angle ACB is greater than

the angle ABC, then will the side AB, opposite the former, be greater than the side AC opposite the latter.

For, if AB is not greater than AC, it must be either equal to it, or less than it. But it cannot be equal; for then the angle ACB would be equal to the angle ABC, (Prop. v,) which, by supposition, is not the case. Neither can it be less; for then the angle ACB would be less than the angle ABC, by the first part of this proposition; which is also contrary to the supposition. Hence, since the side AB is neither equal to AC, nor less than it, it must of necessity be greater.

PROPOSITION XVI.

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

In the triangle ABC, the sum of the two sides AC and BC is greater than the third side AB.

Α

D

C

B

For, produce AC till CD be equal to CB, or AD equal to the sum of the two AC, CB: join BD. Then because CB is equal to CD, the angle D is equal to CBD, (Prop. v.) But the angle ABD is greater than the angle CBD; consequently it must be greater than D. And since the greater side of any triangle is opposite the greater angle, (Prop. xv,) the side AD of the triangle ABD is greater than the side AB. But AD is equal to the sum of AC, CD, or to the sum of AC, CB; therefore AC+CB is greater than AB.

They supposed an ass

(38.) This is the celebrated proposition which the Epicureans derided, as being manifest even to asses. would know that it was further around the across the same corner.

(39.) Let DB be a candlestick with a candle, whose flame is at B, standing perpendicularly on a plane mirror CD. Let A be the position of the eye at the height CA above the mirror. Required to prove that the light, passing from the candle to the mirror, and thence to the eye, obeying the law of nature, that is, making the an

corner of a field, than

B

gle of reflection equal to the angle of incidence, takes the minimum route. In other words, prove that the sum of the two lines BG, GA is less than the sum of any other two lines drawn from B and A to meet in a point of the mirror, such as the two lines BK, KA.

Produce BD until DF is equal to BD, and join KF, and then as in (29,) we can show that KF is equal to KB, so that AK+KB is equal to AK+KF; also AG+GB is equal to AG+GF, or to AF. Now, by the above proposition, AF is less than AK+KF; therefore AG+GB is less than AK+KB.

The time required for the light to pass from B to A, by the reflection of the mirror CD, is, by the law of nature, the shortest possible.

PROPOSITION XVII.

THEOREM.

When a line intersects two parallel lines, it makes the alternate angles equal to each other.

Let the line FG cut the two parallel lines AB, CD; then will the angle AFG be equal to the alternate angle DGF.

For, if they are not equal, one of them must be greater

A

F

D

G

than the other: let it be DGF, for instance, which is the greater, if possible; and conceive the line GB to be drawn, cutting off the part or angle FGB equal to the angle AFG, and meeting the line AB in the point B. The angle AFG, being exterior in reference to the triangle GFB, is greater than the interior and opposite angle FGB, (Prop. VI.) Hence, the angle AFG is greater and equal to FGB, at the same time, which is impossible. Therefore the angle FGD is not unequal to the alternate angle AFG; that is, they are equal to each other.

Cor. Straight lines which are perpendicular to one of two parallel lines, are also perpendicular to the other.

PROPOSITION XVIII.

THEOREM. When a line, cutting two other lines, makes the alternate angles equal to each other, those two lines are parallel.

Let the line FG, cutting the two lines AB, CD, make the alternate angles AFG, DGF, equal to each other; then will AB be parallel to CD.

For if they are not parallel,

F

B

D

H

let some other line, as GH, be parallel to AB. Then, because of these parallels, the angle AFG is equal to the alternate angle HGF, (Prop. xvII;) but the angle AFG is equal to the angle DGF; therefore the angle DGF is equal to the angle HGF, (Ax. I ;) that is, a part is equal to the whole, which is impossible. Therefore no line

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