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ssrs barcode font 4.2B Distance Principles in ObjectiveC
4.2B Distance Principles Denso QR Bar Code Scanner In ObjectiveC Using Barcode Control SDK for iPhone Control to generate, create, read, scan barcode image in iPhone applications. Encoding QR Code In ObjectiveC Using Barcode encoder for iPhone Control to generate, create QR Code 2d barcode image in iPhone applications. PRINCIPLE 1: QR Code Reader In ObjectiveC Using Barcode scanner for iPhone Control to read, scan read, scan image in iPhone applications. Draw Barcode In ObjectiveC Using Barcode encoder for iPhone Control to generate, create bar code image in iPhone applications. If a point is on the perpendicular bisector of a line segment, then it is equidistant from the ends of the line segment. QR Generation In Visual C#.NET Using Barcode creation for .NET Control to generate, create QRCode image in .NET framework applications. Generate Quick Response Code In Visual Studio .NET Using Barcode creation for ASP.NET Control to generate, create QR Code ISO/IEC18004 image in ASP.NET applications. Thus if P is on CD, the ' bisector of AB in Fig. 424, then PA > PB.
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Code 39 Full ASCII Encoder In None Using Barcode drawer for Online Control to generate, create Code 3/9 image in Online applications. Code128 Reader In C#.NET Using Barcode recognizer for VS .NET Control to read, scan read, scan image in Visual Studio .NET applications. If a point is on the bisector of an angle, then it is equidistant from the sides of the angle.
Thus if P is on AB, the bisector of j A in Fig. 425, then PQ > PR, where PQ and PR are the distances of P from the sides of the angle. Fig. 425 PRINCIPLE 4: If a point is equidistant from the sides of an angle, then it is on the bisector of the angle. (Principle 4 is the converse of Principle 3.) Thus if PQ PR, where PQ and PR are the distances of P from the sides of j A in Fig. 425, then P is on AB, the bisector of jA. PRINCIPLE 5: Two points each equidistant from the ends of a line segment determine the perpendicular bisector of the line segment. (The line joining the vertices of two isosceles triangles having a common base is the perpendicular bisector of the base.) Thus if PA > PB and QA > QB in Fig. 426, then P and Q determine CD, the ' bisector of AB.
Fig. 426 PRINCIPLE 6: The perpendicular bisectors of the sides of a triangle meet in a point which is equidistant from the vertices of the triangle. Thus if P is the intersection of the ' bisectors of the sides of ^ ABC in Fig. 427, then PA > PB > PC. P is the center of the circumscribed circle and is called the circumcenter of ^ABC. Fig. 427 CHAPTER 4 Parallel Lines, Distances, and Angle Sums
PRINCIPLE
The bisectors of the angles of a triangle meet in a point which is equidistant from the sides of the triangle. Thus if Q is the intersection of the bisectors of the angles of ^ ABC in Fig. 428, then QR > QS > QT, where these are the distances from Q to the sides of ^ABC. Q is the center of the inscribed circle and is called the incenter of ^ABC. Fig. 428 SOLVED PROBLEMS
Finding distances In each of the following, find the distance and indicate the kind of distance involved. In4 429(a), Fig. 4 4 4 find the distance (a) from P to A; (b) from P to CD; (c) from A to BC; (d ) from AB to CD. In Fig. 429(b), find the distance (e) from P to inner circle O; (f) from P to outer circle O; (g) between the concentric circles. Fig. 429 Solutions
(a) PA (b) PG (c) AE (d) FG (e) PQ (f) PR (g) QR 7, distance between two points 4, distance from a point to a line 10, distance from a point to a line 6, distance between two parallel lines 12 3 12 5 5 3 9, distance from a point to a circle 7, distance from a point to a circle 2, distance between two concentric circles Locating a point satisfying given conditions In Fig. 430. (a) Locate P, a point on BC and equidistant from A and C. (b) Locate Q, a point on AB and equidistant from BC and AC. (c) Locate R, the center of the circle circumscribed about ^ ABC. (d) Locate S, the center of the circle inscribed in ^ABC. Fig. 430 CHAPTER 4 Parallel Lines, Distances, and Angle Sums
Solutions
See Fig. 431. Fig. 431 Applying principles 2 and 4 For each ^ABC in Fig. 432, describe P, Q, and R as equidistant points, and locate each on a bisector. Fig. 432 Solutions
(a) Since P is equidistant from A and B, it is on the ' bisector of AB. Since Q is equidistant from B and C, it is on the ' bisector of BC. Since R is equidistant from A, B, and C, it is on the ' bisectors of AB, BC, and AC. (b) Since P is equidistant from AB and BC, it is on the bisector of jB. Since Q is equidistant from AC and BC , 4 4 4 it is on the bisector of jC. Since R is equidistant from AB , BC, and AC, it is on the bisectors of jA, j B, and jC. 4 4 4 4 Applying principles 1, 3, 6, and 7 For each ^ABC in Fig. 433, describe P, Q, and R as equidistant points. Also, describe R as the center of a circle. Fig. 433

