Dem.—Join CF, CG. Then the two triangles FHC, GHC have FH equal to GH (const.), and HC common; and the base CF equal to the base CG, being radii of the circle FDG (Def. xxxii.). Therefore the angle CHF is equal to the angle CHG [viii.], and, being adjacent angles, they are right angles (Def. xiii.). Therefore CH is perpendicular to AB. Exercises. 1. Prove that the circle cannot meet AB in more than two points. 2. If one angle of a triangle be equal to the sum of the other two, the triangle can be divided into the sum of two isosceles triangles, and the base is equal to twice the line from its middle point to the opposite angle. PROP. XIII.—Theorem. The adjacent angles (ABC, ABD) which one right line (AB) standing on another (CD) makes with it are either both right angles, or their sum is equal to two right angles. Dem.—If AB is perpendicular to CD, as in fig. 1, the angles ABC, ABD are right angles. If not, draw BE perpendicular to CD [xi.]. Now the angle CBA is equal to the sum of the two angles CBE, EBA (Def. xi.). Hence, adding the angle ABD, the sum of the angles CBA, ABD is equal to the sum of the three angles CBE, EBA, ABD. In like manner, the sum of the angles CBE, EBD is equal to the sum of the three angles CBE, EBA, ABD. And things which are equal to the same are equal to one another. Therefore the sum of the angles CBA, ABD is equal to the sum of the angles CBE, EBD; but CBE, EBD are right angles; therefore the sum of the angles CBA, ABD is two right angles. Or thus: Denote the angle EBA by θ; then evidently the angle CBA = right angle + θ; the angle ABD = right angle − θ; therefore CBA + ABD = two right angles. Cor. 1.—The sum of two supplemental angles is two right angles. Cor. 2.—Two right lines cannot have a common segment. Cor. 3.—The bisector of any angle bisects the corresponding re-entrant angle. Cor. 4.—The bisectors of two supplemental angles are at right angles to each other. Cor. 5.—The angle EBA is half the difference of the angles CBA, ABD. PROP. XIV.–Theorem. If at a point (B) in a right line (BA) two other right lines (CB, BD) on opposite sides make the adjacent angles (CBA, ABD) together equal to two right angles, these two right lines form one continuous line. Dem.—If BD be not the continuation of CB, let BE be its continuation. Now, since CBE is a right line, and BA stands on it, the sum of the angles CBA, ABE is two right angles (xiii.); and the sum of the angles CBA, ABD is two right angles (hyp.); therefore the sum of the angles CBA, ABE is equal to the sum of the angles CBA, ABD. Reject the angle CBA, which is common, and we have the angle ABE equal to the angle ABD—that is, a part equal to the whole—which is absurd. Hence BD must be in the same right line with CB. PROP. XV.—Theorem. If two right lines (AB, CD) intersect one another, the opposite angles are equal (CEA = DEB, and BEC = AED).