Cor. The perpendicular drawn from the middle of the base. of an isosceles triangle passes through the vertex and bisects the vertical angle.

PROPOSITION XXV. THEOR.

Of all the straight lines that can be drawn from the same point to a straight line, the perpendicular is the least; of the others that which is nearer the perpendicular is less than those more remote, and from that point to the given line two equal straight lines only can be drawn, one on each side of the least, and at an equal distance from it.

From the point C draw CD perpendicular to AB, take E and F, any points on one side CD, and join CE and CF. The perpendicular CD is the least. For in the triangle ECD the angle EDC is right, and therefore (17, Cor.) the angle CED is acute, and less than EDC; and the less side is (18, Cor.) opposite to the less angle, therefore CD is less than CE. Again, CE is less than CF; for CEF being the exterior angle of the triangle CED, is greater than the interior opposite angle CDE; therefore the angle CEF is obtuse; and therefore any other angle, CFE of the triangle CFE, must be (17, Cor.) acute, and less than CEF, and the less side is opposite to the less angle; hence CE is less than CF.

AF E D G B

Make (2) DG equal to DE, and join CG. In the triangles CDE and CDG, the side ED is equal to DG, and CD is common to both, and the included angles at D are equal; therefore (12) the triangles are equal, and CG is equal CE; also CG is the only line that can be drawn equal to CE; for if any other line were drawn, it must lie either on the one side of CG or the other, and hence would be either greater or less than it, and therefore unequal to CE.

Cor. 1. The perpendicular of a triangle must always be less than its sides, and the angle or angles opposite to it acute; therefore if the two angles at the base of a triangle be acute, the perpendicular must fall within the triangle; if one of them be obtuse, the perpendicular must fall without it, on the side of the obtuse angle.

Cor. 2. The greater side of every triangle is greater than any straight line that can be drawn from the vertex to the base within the triangle.

PROPOSITION XXVI. THEOR.

If two right-angled triangles have the hypotenuse and a side in one equal to the hypotenuse and a like side in the other, the triangles are equal.

Let ABC and DEF be two right-angled triangles which have the hypotenuse BA, and the side AC in the one equal to the hypotenuse ED, and the side DF in the other, the triangles are equal.

A

44

B

C E

F

For if the triangle ABC be applied to the triangle DEF, so that the point B may be on E, and the straight line BA on ED, the point A shall coincide with D, because BA is equal ED; and since there can be only one perpendicular from D to the base EF (25), AC must coincide with DF, and the point C shall coincide with F, because AC is equal to DF; but the point B coincides with E; therefore (Def. 3, Cor. 1) the base BC shall coincide with EF, and be equal to it; wherefore the two plane surfaces ABC and DEF have equal boundaries, and coinciding in three points not in the same straight line, they must coincide through their whole extent, and be in all respects equal; their areas are equal, the angle A is equal to D, and the angle B to E.

PROPOSITION XXVII. PROB.

At a given point in a given straight line to make an angle equal to a given rectilineal angle.

Let AB be the given straight line, and A a given point in it, and DCE the given rectilineal angle, it is required to make an angle at the point A that shall be equal to the given angle DCE.

Take in CD and CE any points D and E, and join DE, and construct the triangle AFG, having its three sides (10) respectively equal to the three given straight lines CD, DE, EC; and because the triangles CDE and AFG

have the three sides of the one re

A

E

G

F

spectively equal to the three sides of the other, they are equal in every respect, and therefore the angle FAG is equal to the angle DCE.

If two triangles have two sides of the one equal to two sides of the other, each to each, but the angle included by the two sides of the one greater than the angle included by the two sides of the other, the base of that which has the greater angle is greater than the base of the other.

Let ABC and DEF be two triangles which have the two sides AB, AC equal to the two DE, DF, each to each, but the angle BAC greater than the A.

angle EDF, the base BC is also

greater than the base EF.

B

H

E

C

F

Of the two sides DE, DF, let DE be the side which is not greater than the other, and at the point D, in the straight line DE, make (27) the angle EDG equal to the

D

angle BAC, and make (2) DG equal to AC, or DF, and join EG, GF. Because DG is greater than DE, it is also greater (25, Cor. 2) than the line DH drawn from the vertex to the base within the triangle DEG, but DG is equal to DF; therefore DF is greater than DH, and the point F must lie below the line EG. Because the side AB is equal to DE, and the side AC to DG, and the included angle BAC is equal to the included angle EDG; therefore (12) the triangle ABC is equal to DEG, and the base BC equal to the base EG; and because DG is equal to DF, the angle DFG is equal (14) to DGF, but the angle DGF is greater than EGF; therefore the angle DFG is greater than EGF; much more is the angle EFG greater than EGF; but the greater side of every triangle (20) is opposite to the greater angle; therefore the side EG of the triangle EGF is greater than its side EF, but EG is equal to BC; therefore BC is also greater than EF.

PROPOSITION XXIX. THEOR.

If two triangles have two sides of the one equal to two sides of the other, each to each, but the base of the one greater than the base of the other, the angle included by the sides of that which has the greater base is greater than the angle included by the sides of the other.

Let ABC, DEF be two triangles, of which the side AB is equal to DE, and the side AC to DF, but let the base BC be greater than the base EF, the angle BAC is also greater than the angle EDF.

For if it be not greater it must either be equal to it or less; but the angle BAC is not equal to EDF, because then the base BC would be (12) equal to the base EF, but it is not; therefore the angle BAC is

F

not equal to EDF, neither is it less, because then the base BC

would be less (28) than the base EF, but it is not; therefore the angle BAC is not less than EDF, and it was shown not to be equal to it; therefore the angle BAC is greater than the angle EDF.

PROPOSITION XXX. THEOR.

The three interior angles of every triangle are together equal to two right angles.

Let ABC be a triangle, its three interior angles A, B, and C, are together equal to two right angles.

B

A

From the point A draw any straight line AD to the base. Suppose the point A to move along the line AD, while the sides vary to suit its progress, it is evident that the form of the triangle must vary with every change of the position of the point A. Let the point A have moved to E and join EB and EC. On comparing the triangle BEC with the original triangle, it appears that the sides and the angles at the base have become (21) less, and the vertical angle greater, than those of the original triangle. Suppose now the point A to have moved on till it coincide with D, then the sides become equal to BD and DC; the angles at the base disappear, the vertical angle becomes equal (Ax. 9) to the three angles of the triangle, and at the same time equal to the two angles ADB and ADC, which are (5) together equal to two right angles; wherefore the three interior angles of every triangle are together equal to two right angles.

Cor. 1. If two angles of one triangle be equal to two angles. of another, the triangles are equiangular.

Cor. 2. If one angle of a triangle be a right angle, the other two are together equal to a right angle.

Cor. 3. Any angle of an equilateral triangle is one-third of two right angles, or two thirds of one right angle.