Let ABC, DEF be two triangles, in which the side AB is equal to the side DE, and the side AC equal to the side DF, and the included angle BAC equal to the included angle EDF, then shall the base BC be equal to the base EF, and the triangle ABC to the triangle DEF, and the other angles, each to each, to which the equal sides are opposite, namely, the angle ABC to the angle DEF, Α Demons Demonstration. Let the triangle ABC be applied to the triangle DEF, so that the point A may be upon the point D, and the straight line AB upon the straight line DE;. then, since AB is equal to DE, Again, since AB coincides with DE, therefore AC will fall upon DF. Now, B coinciding with E, and C with F, the straight line BC will coincide with the straight line EF, for if it does not, it must take some other position as EGF, in which case two straight lines would enclose a space, which is impossible: (ax. 10) therefore BC does coincide with EF, and is equal to it; (ax. 8) and the triangle ABC is equal to the triangle DEF; (ax. 8) and the other angles are equal, namely, ABC to DEF, and ACB to DFE. Q.E.D. Exercises. 1. State what parts of the triangles are given equal in the fourth proposition, and shew how each of these data is used in the proof. 2. A straight line A B is drawn at right angles to another straight line CD from its middle point B. Shew that AC and AD are of equal length. 3: The vertical angle BAC of an isosceles triangle ABC is bisected (iie divided into two equal parts) by a straight line AD (D being outside the line BC); shew that DBC is an isosceles triangle. 4. Two straight lines are drawn bisecting each other at right angles; shew that any point in either of them is equidistant from the extremities of the other. 5. If the straight line which joins the vertex of a triangle with the middle point of its base be at right angles to the base, the triangle is isosceles. 6. A straight line is drawn bisecting the vertical angle of an isosceles triangle ; shew that it also bisects the base, and at right angles. 7. From the sides AB, AC of an isosceles triangle ABC, equal parts AD, AE are cut off ; shew that BE is equal to CD. 8. The triangle formed by joining the middle points of the sides of an equilateral triangle is also equilateral. 9. ABCD is a quadrilateral such that AB is equal to BC, and BD bisects the angle ABC; shew that AD is equal to CD, and that BD bisects the quadrilateral and the angle ADC. PROPOSITION V. THEOREM. The angles at the base of an isosceles triangle are equal to each other. Let ABC be an isosceles triangle, in which the side AB is equal to the side AC; then shall the angle ABC be equal to the angle ACB. CB Demonstration. Let the triangle ABC be taken up, turned round, and put down again with the position of its sides reversed; let AB in its new position be named AB', AB is equal to AC', AC is equal to AB', and the angle BAC is equal to the angle C'AB'; therefore the angle ABC is equal to the angle AC'B', (prop. 4) but the angle AC'B' is the same as the angle ACB, therefore the angle ABC is equal to the angle ACB. Q.E.D. COROLLARY.-Hence every equilateral triangle is also equiangular, PROPOSITION V. [NOTE.-The following is Euclid's more complete statement and proof of this proposition.] The angles at the base of an isosceles triangle are equal to each other; and if the equal sides be produced, the angles on the other side of the base shall be equal. Let ABC be an isosceles triangle in which the side AB is equal to the side AC, Construction. In BD take any point F; from AE the greater cut off AG equal to AF the less ; (prop. 3.) join FC, GB. Demonstration. and AC to AB; (hyp.) and the triangle AFC is equal to the triangle AGB, and the remaining angles of the one to the remaining angles of the other, each to each, to which the equal sides are opposite, namely, the angle ACF to the angle ABG, and the angle AFC to the angle AGB. Again, since the whole AF is equal to the whole AG, (constr.) and parts of these AB, AC are equal, (hyp.) therefore the remainder BF is equal to the remainder CG; (ax. 3) also it has been shewn that FC is equal to GB : hence in the triangles BFC, CGB, the two sides BF, FC are equal to the two sides CG, GB, each to each, and the included angle BFC has been shewn equal to the included angle CGB; therefore these triangles are equal in all respects, (prop. 4) and the angle FBC is equal to the angle GCB, Now it has been demonstrated that the whole angle ABG is equal to the whole angle ACF, and that the parts of these CBG, BCF are also equal, therefore the remaining angle ABC is equal to the remaining angle ACB, (ax. 3) which are the angles at the base of the triangle ABC. It has also been shewn that the angle FBC is equal to the angle GCB, which are the angles on the other side of the base. Q.E.D. COROLLARY.—Hence every equilateral triangle is also equiangular. Exercises. 1. ABC, DBC are two isosceles triangles on the same base and on the same side of it; prove that A D bisects the angle BAC. 2. If two isosceles triangles stand upon opposite sides of the same base, the straight line which joins their vertices will bisect the vertical angles ; and also will bisect the base at right angles. 3. In the figure on page 14, shew that the angle DBG is equal to the angle ECF. 4. If a point G be found in an equilateral triangle so that GA, GB, GC are all equal, then GA bisects the angle A : also if points E, F be taken in AC, A B so that AE is equal to AF, then GE is equal to GF. 5. The straight lines drawn from the angular points of an equilateral triangle to the middle points of the opposite sides are equal to each other. PROPOSITION VI. THEOREM. If the angles at the base of a triangle be equal to each other, the triangle is isosceles. Let ABC be a triangle, in which the angle B is equal to the angle C; then shall AB be equal to AC. Demonstration. again with its sides reversed ; and C in its new position be named C". then B' will coincide with C, because C'B' is equal to BC. the angles B and C" are also equal, similarly B'A will fall upon CA : Q.E.D. COROLLARY.—Hence every equiangular triangle is also equilateral. PROPOSITION VI. THEOREM. [The following is Euclid's proof of this proposition.] If two angles of a triangle be equal to one another, the sides also which subtend (that is, are opposite to) the equal angles shall be equal to one another. |