EXERCISES 210. AD and BE are perpendiculars from the vertices of a triangle ABC to the opposite sides. Prove that AC:DC = BC:EC. : 211. ABC is a straight line and ABD and BCE are triangles on the same side of it, such that AB: BE = BC :BD and the angle DBA equals the angle EBC. If AE and DC intersect at 0, prove that AOC is an isosceles triangle. 212. A square is inscribed in a right triangle, having one of its sides coincident with a segment of the hypotenuse. Prove that a side of the square is a mean proportional between the other two segments of the hypotenuse. 213. The quadrilateral ABCD is inscribed in a circle. CB and DA produced meet at E. A bisector of the angle E meets AB at F, and CD at G. Prove that BA:BF= DC: DG. 214. If two circles touch each other, straight lines drawn through the point of tangency are cut proportionally by the circumferences. 215. Two secants are drawn to a circle from an outside point and their external segments are 20 and 24. If the internal segment of the former is 16, what is the internal segment of the latter ? 216. Tangents to two intersecting circles from any point in their common chord produced are equal. 217. If two circles intersect, their common chord produced bisects their common tangents. 218. If, in a right triangle, a perpendicular is drawn from the vertex of the right angle to the hypotenuse, and circles inscribed in the two triangles formed, their radii are proportional to the legs of the right triangle. 219. ABC is a triangle inscribed in a circle, and D is the middle point of the arc AB. Prove that CD' = AC X BC + AD'. 220. AB is a diameter of a circle, and tangents are drawn at A and B. A tangent to the circle at any other point C meets the two first tangents at E and F. O is the centre of the O. Prove EOP: FOʻ = CE: CF. = a 221. ABC is an equilateral triangle inscribed in a circle. Lines are drawn to the three vertices from any point in the circumference, as, from D in the arc AC. Prove that DB=DA+DC. (Hint: draw chord AE parallel to DC.) 222. P is a point in AC, the diagonal of a parallelogram ABCD. EF is drawn through P, meeting BC and AD at E and F respectively, and GH is drawn through P, meeting AB and CD at G and H respectively. Prove EH parallel to GF. 223. In any triangle, the product of any two sides is equal to the product of the segments of the third side formed by a bisector of the exterior angle at the opposite vertex, minus the square of the bisector. Hint. Circumscribe a circle and produce the diameter to the circumference as in § 223. 224. If the sides of a triangle are 8, 10, and 12, find the length of the bisector of the exterior angle formed by producing either of the first two sides ($ 202, and the preceding exercise). 225. A chord of a circle CD intersects a diameter AB at right angles, and E is any point in CD. AE and BE produced meet the circumference at F and G respectively. Prove that CG:CF = DG: DF. 226. D is the middle point of the side BC of the triangle ABC. CE is any line intersecting AB at E and AD at F. Prove that the ratio of AF to FD is twice the ratio of AE: EB. 227. BC is the hypotenuse of a right triangle ABC, and AD a median. AE is drawn to the hypotenuse, making ABE an isosceles triangle. Prove that BE:BA = BA: BD. 228. O is the centre of the circle inscribed in the triangle ABC and AO produced meets BC at D. Prove that OA: OD = AB + AC: BC. 229. AB and CD intersect at E, and AC is parallel to DB. If F and G are points in AC and DB such that FA: FC = GB: GD, prove that F, E, and G lie in a straight line. BOOK IV AREAS — MEASUREMENT OF THE CIRCLE DEFINITIONS 235. The Area of a polygon is the measure of its surface, i.e. the numerical value of its ratio to a unit of surface, arbitrarily taken as a standard. A unit of surface is usually a square whose side is some unit of length, as a square foot, or a square decimeter. 236. Equivalent Polygons are those which have equal areas. 237. The Altitude of a triangle, as previously stated, is the perpendicular distance from the opposite vertex to the base, or to the base produced, as a. 238. The Altitude of a parallelogram or a trapezoid is the perpendicular distance between the two bases, as a. a ADA a a The words triangle, quadrilateral, etc., are hereafter often used meaning the area of the triangle, area of the quadrilateral, etc. 121 PROPOSITION I 239. Theorem. Two polygons mutually equiangular and mutually equilateral are equal. 240. Theorem. Two rectangles having equal altitudes are to each other as their bases. Cons. Apply com. meas., such as AH, to AD and AF, and let it be contained m times in AD, and n times in AF = |