The above construction will also apply in case the ground is horizontal, and the trees, instead of being vertical, are oblique.

(32.) The foregoing are only particular cases of the following more general proposition:

PROBLEM. To find a point in a given straight line, equally distant from two given points.

(33.) Suppose AC to be a tree, standing on the horizontal plane AB; it is required to find at what point it must be broken, so that, by falling, the top may strike the ground at B.

A

D

B

Solution. Join CB, and bisect it by the perpendicular DF; then will F be the point at which the tree must break. For, joining BF, and comparing the triangle FBD with the triangle FCD, we see that the side DB is equal to DC, the side FD common, and the contained angle FDB equal to FDC, each being a right-angle. Therefore (Prop. 11,) the triangle FBD is equal to FCD: consequently FB is equal to FC.

The solution is effected in the same way when the tree is supposed to stand upon an inclined surface, as upon the side

of a hill.

A A

B

(34.) The general proposition, of which the foregoing are particular cases, may be thus given:

PROBLEM. Given the base of a triangle, one of the angles at the base, and the sum of the other two sides, to construct the triangle.

25

Construction. Make AB equal to the given base; draw AC, making the angle BAC equal to the given angle, and AC equal to the sum of the other two sides; join BC, and bisect it by the perpendicular DF; finally, join BF, and ABF will be the triangle required. This is obvious from what has already been done.

PROPOSITION IV.

THEOREM. If two triangles have two angles and the interjacent side of the one equal to two angles and the interjacent side of the other, the triangles will be identical, or equal in all respects.

In the two triangles ABC, DFG, if the angle A is equal to the angle D, the angle B equal to the angle F, and the side AB equal to the side DF; then will the triangles be identical, or equal in all respects.

For, conceive the triangle ABC to be placed on the triangle DFC, in such a manner that the side AB may coincide with the equal side DF. Then, since the angle D is equal to the angle A, the side AC will take the direction of the side DG; also, since the angle F is equal to the angle B, the side BC will take the direction of the side FG; consequently the point C must coincide with

the point G. Therefore the two triangles are identical. (Ax. IX,) having the two sides AC and BC respectively equal to DG and FG, and the remaining angle C equal to the remaining angle G.

THEOREM.

PROPOSITION V.

In an isosceles triangle, the angles at the base are equal; or, if a triangle have two sides equal, the angles opposite those sides will be equal.

If the triangle ABC have the side AC equal to the side BC, then will the angle B be equal to the angle A.

For, conceive the angle C to be bisected, or divided into two equal parts, by the line CD, making the angle ACD equal to the angle BCD. Then the two triangles ADC and BDC have two sides, and the included angle of the one equal to the two sides and the included angle of the other, namely, the side AC equal to BC, the angle ACD equal to the angle BCD; and the side CD common; therefore the two triangles are identical, or equal in all respects, (Prop. 111,) and consequently the angle B is equal to the angle A.

Cor. 1. Hence the line which bisects the vertical angle of an isosceles triangle, bisects the base, and is also perpendicular to it.

Cor. 2. It also appears that every equilateral triangle is equiangular, or has all its angles equal.

PROPOSITION VI.

THEOREM. When one side of a triangle is produced, the exterior angle is greater than either of the two interior and opposite angles.

Let ABC be a triangle, having the side AB produced to D; then will the exterior angle CBD be greater than either of the interior and opposite angles BAC or BCA.

A

B

H

For, conceive the side BC to be bisected in the point F, and draw the line AF and produce it until FG is equal to AF; and join BG. Now, in the two triangles AFC and GFB, the sides FA and FC are respectively equal to the sides FG and FB, and the opposite angles AFC and GFB are equal; (Prop. II ;) therefore these two triangles are equal in all respects, (Prop. III,) and we have the angle ACF equal to the angle GBF; consequently the exterior angle CBD, being greater than GBF, is greater than the interior angle BCA.

In like manner, if CB be produced to H, and AB be bisected, it may be shown that the exterior angle ABH, or its equal CBD, is greater than the other interior angle BAC.

PROPOSITION VII.

THEOREM. When a triangle has two of its angles equal, the sides opposite to them are also equal, and the triangle will be isosceles.

In the triangle ABC, let the angle CAB be equal to the angle CBA; then will the side CB be equal to CA.

For, if these sides are not equal, one must be greater than the other let CA be greater than CB; then take AD equal to BC, and join DB. Now, in the two triangles DAB and CBA, we have DA equal to CB by construction, the side AB common, and the angle CAB equal to the angle CBA by hypothesis; therefore two sides and the included angle of the one are respectively equal to two sides and the included angle of the other; consequently the triangle DAB is equal to the triangle CBA, (Prop. III.) But a part cannot be equal to the whole; (Ax. VIII;) hence there can be no inequality between the sides CB and CA, and therefore the triangle is isosceles.