Dem.—Because the line AE stands on CD, the sum of the angles CEA, AED is two right angles [xiii.]; and because the line CE stands on AB, the sum of the angles BEC, CEA is two right angles; therefore the sum of the angles CEA, AED is equal to the sum of the angles BEC, CEA. Reject the angle CEA, which is common, and we have the angle AED equal to BEC. In like manner, the angle CEA is equal to DEB. The foregoing proof may be briefly given, by saying that opposite angles are equal because they have a common supplement. Questions for Examination on Props. XIII., XIV., XV. 1. What problem is required in Euclid’s proof of Prop. xiii.? 2. What theorem? Ans. No theorem, only the axioms. 3. If two lines intersect, how many pairs of supplemental angles do they make? 4. What relation does Prop. xiv. bear to Prop. xiii.? 5. What three lines in Prop. xiv. are concurrent? 6. What caution is required in the enunciation of Prop. xiv.? 7. State the converse of Prop. xv. Prove it. 8. What is the subject of Props. xiii., xiv., xv.? Ans. Angles at a point. PROP. XVI.—Theorem. If any side (BC) of a triangle (ABC) be produced, the exterior angle (ACD) is greater than either of the interior non-adjacent angles. Dem.—Bisect AC in E [x.]. Join BE (Post. i.). Produce it, and from the produced part cut off EF equal to BE [iii]. Join CF. Now because EC is equal to EA (const.), and EF is equal to EB, the triangles CEF, AEB have the sides CE, EF in one equal to the sides AE, EB in the other; and the angle CEF equal to AEB [xv.]. Therefore [iv.] the angle ECF is equal to EAB; but the angle ACD is greater than ECF; therefore the angle ACD is greater than EAB. In like manner it may be shown, if the side AC be produced, that the exterior angle BCG is greater than the angle ABC; but BCG is equal to ACD [xv.]. Hence ACD is greater than ABC. Therefore ACD is greater than either of the interior non-adjacent angles A or B of the triangle ABC. Cor. 1.—The sum of the three interior angles of the triangle BCF is equal to the sum of the three interior angles of the triangle ABC. Cor. 2.—The area of BCF is equal to the area of ABC. Cor. 3.—The lines BA and CF, if produced, cannot meet at any finite distance. For, if they met at any finite point X, the triangle CAX would have an exterior angle BAC equal to the interior angle ACX. PROP. XVII.—Theorem. Any two angles (B, C) of a triangle (ABC) are together less than two right angles. Dem.—Produce BC to D; then the exterior angle ACD is greater than ABC [xvi.]: to each add the angle ACB, and we have the sum of the angles ACD, ACB greater than the sum of the angles ABC, ACB; but the sum of the angles ACD, ACB is two right angles [xiii.]. Therefore the sum of the angles ABC, ACB is less than two right angles. In like manner we may show that the sum of the angles A, B, or of the angles A, C, is less than two right angles.