PROP. 30.-Theorem.-Straight lines which are parallel to the same straight line are parallel to one another. Let the straight lines AB, CD be each of them parallel to PQ; then AB shall be parallel to CD. Construction -Draw the straight line EF, cutting AB, CD and Proof. Because AB is parallel to PQ, and EF falls on them, therefore the angle AGH is equal to the alternate angle GKQ. Prop. 29 Again, because CD is parallel to PQ, and EF falls on them, therefore the angle GHD is equal to its interior opposite angle GKQ, Prop. 29 therefore the angle AGH is equal to the angle GHD; Ax. 1. but these are alternate angles ; therefore AB is parallel to CD. Prop. 27. Q. E. D. EXERCISE. The two lines AB and CD are each perpendicular to the same line BD; prove that AB and CD are parallel. PROP. 31.-Problem.-Through a given point to draw a straight line parallel to a given straight line. Let A be the given point and BC the given straight line; it is required to draw a straight line through A, parallel to BC. At the point A in the straight line AD, make the angle EAD equal to the alternate angle ADC. Produce EA to F. Then EF shall be parallel to BC. Prop. 23 Proof. Because the straight line AD falls on the two straight lines EF, BC, and makes the angle EAD equal to its alternate angle ADC, therefore EF is parallel to BC, and it has been drawn through the given point A. Prop. 27 EXERCISES. 1. ABC is an isosceles triangle, having the side AB equal to the side AC; in AB take any point P and draw PQ parallel to BC; show that the angle APQ is equal to the angle AQP. 2. In the figure of Prop. 31, suppose EA is equal to DC; join ED and AC; prove that the triangles EAD, CDA are equal in all respects. 3. Prove the following alternative construction for Prop. 31:-Take any points, D, E in BC, and join AD; with centre A, and radius equal to DE, describe a circle; and with centre E and radius equal to AD, describe another circle, cutting the former circle in F and G. If F and D are on opposite sides of AE, AF will be parallel to BC. EUC. F PROP. 32.-Theorem.—If one side of a triangle be produced, the exterior angle is equal to the sum of the two interior opposite angles; and the three angles of any triangle are together equal to two right angles. Let ABC be a triangle, and let the side BC be produced to D; (1) the exterior angle ACD shall be equal to the sum of the two interior opposite angles BAC and ABC; and (2) the three angles BAC, ABC and ACB of the triangle ABC shall be together equal to two right angles. E D Construction. Through C draw CE parallel to AB. Prop. 31 Proof. (1) Because CE is parallel to AB, and AC falls on them, therefore the angle ACE is equal to the alternate angle BAC. Pp. 29 Again, because CE is parallel to AB, and BCD falls on them, therefore the exterior angle ECD is equal to its interior opposite angle ABC. Prop. 29 Therefore the two angles ACE, ECD, that is, the angle ACD is equal to the sum of the two angles BAC, ABC. Q. E. D. (2) Because the angle ACD is equal to the sum of the angles Proved BAC, ABC, add to each of these equals the angle ACB, therefore the two angles, ACD, ACB are together equal to the three angles BAC, ABC and ACB. But the angles ACD, ACB are together equal to two right angles. Therefore the three angles BAC, ABC and ACB are together equal to two right angles. Prop. 13 Q. E. D. Corollary 1.-All the interior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides. Let ABCDE be any rectilineal figure ; all the interior angles of ABCDE, together with four right angles, shall be equal to twice as many right angles as ABCDE has sides. B E C Construction. Take any point 0 within the figure ABCDE. Join O to each of the angular points A, B, C, D, E. Then ABCDE will be divided into as many triangles as ABCDE has sides. Proof. By Proposition 32 the three angles of any triangle are together equal to two right angles, therefore all the angles of all the triangles in ABCDE are equal to twice as many right angles as ABCDE has sides. But all the angles of all the triangles in ABCDE make up all the interior angles of ABCDE together with the angles at 0, and the angles at 0 are equal to four right angles; Pp. 15, Cor. 2 Therefore all the interior angles of ABCDE together with four right angles are equal to twice as many right angles Q. E. D. EXERCISES. 1. ABC is an equilateral triangle. What is the size of each of its angles? 2. ABC is a triangle having the angle ABC equal to the sum of the angles ACB and BAC. What sort of angle is ABC? 3. ABC is a triangle having each of the angles ABC and ACB double of the angle BAC. What is the size of the angle BAC ? ALTERNATIVE PROOF. Corollary 1.-All the interior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides. Let ABCDE be any rectilineal figure; all the interior angles of ABCDE, together with four right angles, shall be equal to twice as many right angles as ABCDE has sides. A B E Construction.-From A draw as many diagonals, AC, AD, etc., as possible to opposite angles of ABCDE. Then ABCDE will be divided into triangles, and the number of triangles will be less by two than the number of sides of ABCDE. Proof.-By Proposition 32 the three angles of any triangle are together equal to two right angles, therefore all the angles of all the triangles in ABCDE, together with four right angles, are equal to twice as many right angles as ABCDE has sides. But all the angles of all the triangles in ABCDE make up all the interior angles of ABCDE. Therefore all the interior angles of ABCDE, together with four right angles, are equal to twice as many right angles as ABCDE has sides. EXERCISES. Q. E. D. 1. What will the sum of the interior angles be equal to in rectilineal figures of 4, 5, 6, and 7 sides respectively? 2. In the last exercise, if the figures mentioned were equiangular, what would be the size of the angles in each figure? 3. What do you know about-(1) the sum of the two acute angles of a right-angled triangle, and (2) the two acute angles of a right-angled isosceles triangle? |