Plane and solid geometry . P and PBE^ PB=PB. 2. Z DBP = Z PBE, 3. .-. A DBP = APBE, 4. .*. PD = PE, 253. Prop. XL may be stated as follows: Every point in the bisector of an angle is equidistant from thesides of the angle, Ex. 322. Find a point in one side of a triangle which is equidistantfrom the other two sides of the triangle. Ex. 323. Find a point equidistant from two given intersecting linesand also at a given distance from a fixed third line. Ex. 324. Find a point equidistant from two given intersecting linesand also equidistant from two given parallel lines. Ex. 325. Find a poin
Plane and solid geometry . P and PBE^ PB=PB. 2. Z DBP = Z PBE, 3. .-. A DBP = APBE, 4. .*. PD = PE, 253. Prop. XL may be stated as follows: Every point in the bisector of an angle is equidistant from thesides of the angle, Ex. 322. Find a point in one side of a triangle which is equidistantfrom the other two sides of the triangle. Ex. 323. Find a point equidistant from two given intersecting linesand also at a given distance from a fixed third line. Ex. 324. Find a point equidistant from two given intersecting linesand also equidistant from two given parallel lines. Ex. 325. Find a point equidistant from the four sides of a rhombus. Ex. 326. The two altitudes of a rhombus are equal Prove. Ex. 327. Construct the locus of the center of a circle of given radius,which rolls within a given angle so that it always touches a sid6 of theangle. Do not prove. BOOK I 99 Proposition XLL Theorem(Opposite of Prop. XL) 254. The two perpendiculars to the sides of an an^lefrom any point not in its bisector are unequal. C. Given /.ABC\ P any point not in BR, the bisector of AabC\PD and PE, Js from P to BA and BC respectively. To prove PD=f^PE. Outline of Proof Draw FG *^ draw PG. Then FE = FG. Now PF + FG> PG. .*. PE > PG, But PG > PD. •*. PE > PD. 255. Prop. XLI may be stated as follows: Every point not in the bisector of an angle is not equidistantfrom the sides of the angle. 256. Cor. I. (Converse of Prop. XL). Every point equi-distant from the sides of an angle lies in tJie bisector ofthe angle. Hint. Prove directly, using the figure for § 252, or apply § 137. 257. Cor. II. The bisector of an angle is tlxe locus ofall points equidistant from the sides of the angle. Ex. 328. Wliat is the locus of all points that are equidistant froma pair of intersecting lines ? 100 PLANE GEOMETRY CONCUERENT LINE THEOREMSPROPOsiTioisr XLII. Theorem 258. The. bisectors of the angles of a triangle are corvcurrent in a -point which is equidistant from the threesides of the
Size: 1837px × 1359px
Photo credit: © The Reading Room / Alamy / Afripics
License: Licensed
Model Released: No
Keywords: ., bookcentury1900, bookdecade1910, booksubjectgeometr, bookyear1912
