 The measure of an exterior angle of a triangle is equal to the sum of the measures of the two remote interior angles.  Elements of Geometry - Page 79by Andrew Wheeler Phillips, Irving Fisher - 1897 - 354 pagesFull view - About this book
 | Joe Garner Estill - Geometry - 1896 - 168 pages
...AP. PB=A, PL P!Bi. Brown University, June, 1896. 1. Have you been over all the required work ? 2. The exterior angle of a triangle is equal to the sum of the opposite ulterior angles. 3. Find a point equidistant from two given points P and Q, and at a given distance... | |
 | Andrew Wheeler Phillips, Irving Fisher - Geometry, Modern - 1896 - 276 pages
...AC and AE. To PROVE — angle m is measured by \ (arc CE — arc BD). Join CD. Then m + w=s. § 59 [An exterior angle of a triangle is equal to the sum of the two opposite interior angles.] Hence m = s—w. Ax. 3 But s is measured by i arc CE. § 197 And... | |
 | Andrew Wheeler Phillips, Irving Fisher - Geometry - 1897 - 376 pages
...and AE. To PROVE — angle m is measured byl (arc CE — arc BD). Join C and D. Then ;// + K- = J. §58 [An exterior angle of a triangle is equal to the sum of the two opposite interior angles.] Hence in = s—w. Ax. 3 But s is measured by \ arc CE. § 189 And... | |
 | Webster Wells - Geometry - 1898 - 264 pages
...85. Cor. I. It follows from the above demonstration that Z.BCD = Z.ECD + Z.BCE = ZA + ZB; hence 1. An exterior angle of a triangle is equal to the sum of the two opposite interior angles. 2. An exterior angle of a triangle is greater than either of the... | |
 | Education - 1899 - 658 pages
...the angles of a. triangle meet at a point that is equi-distant from the sides of the triangle. 4. The exterior angle of a triangle is equal to the sum of the opposite interior angles. 5. Inscribe a square in a given circle. 6. Give rule for finding area of a regular polygon, and demonstrate... | |
 | George Albert Wentworth - Geometry, Modern - 1899 - 272 pages
...COR. 7. In an equiangular triangle, each angle is one third of two right angles, or 60°. 137. COR. 8. An exterior angle of a triangle is equal to the sum of the two opposite interior angles, and therefore greater than either of them. PROPOSITION XIX. THEOREM.... | |
 | Webster Wells - Geometry - 1899 - 424 pages
...Cor. I. It follows from the above demonstration that Z BCD = Z ECD + Z J3OS = Z .4 + Z 5 ; hence 1. An exterior angle of a triangle is equal to the sum of the two opposite interior angles. 2. An exterior angle of a triangle is greater than either of the... | |
 | Charles Austin Hobbs - Geometry, Plane - 1899 - 266 pages
...COB. IV. In a right triangle, the two acute angles are complementary, Proposition 38. Theorem. 49. An exterior angle of a triangle is equal to the sum of the two opposite interior angles. COB. An exterior angle of a triangle is greater than either of the... | |
 | George Albert Wentworth - Geometry - 1899 - 496 pages
...COR. 7. In an equiangular triangle, each angle is one third of two right angles, or 60°. 137. COR. 8. An exterior angle of a triangle is equal to the sum of the two opposite interior angles, and therefore greater than either of them. PROPOSITION XIX. THEOREM.... | |
 | Arthur Schultze, Frank Louis Sevenoak - Geometry - 1901 - 394 pages
...equal respectively to the hypotenuse and an acute angle of the other. PROPOSITION XIII. THEOREM 100. An exterior angle of a triangle is equal to the sum of the two remote interior angles. B CD Hyp. Z ACD is an exterior angle of A ABC. To prove Z ACD = HIMT.... | |
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