 Every point in the bisector of an angle is equidistant from the sides of the angle. Hyp. Z DAB = Z DAC and 0 is any point in AD. To prove 0 is equidistant from AB and AC. Draw OP _L AB and OP' _L AC, and prove the equality of the two triangles.  Lessons in Geometry: For the Use of Beginners - Page 73by George Anthony Hill - 1888 - 182 pagesFull view - About this book
 | 1903 - 638 pages
...Explain how to clear an equation of fractions and the reason for the method. GEOMETRY. — 1. Prove that every point in the bisector of an angle is equidistant from the sides of the angle. 2. Prove that in the same circle chords equally distant from the center are equal. 3. Show how to circumscribe... | |
 | George Albert Wentworth - 1884 - 264 pages
...from the ends of a straight line determine the perpendicular through the middle point of the line. 69. Theorem. Every point in the bisector of an angle is equidistant from the sides of the angle, and every point not in the bisector is unequally distant from the sides. 70. Theorem. The line joining... | |
 | George Albert Wentworth - Geometry, Analytic - 1886 - 334 pages
...it. 35. The bisectors of the angles contained between the lines y = 2z + 4 and — y = 3x-\-6. HINT. Every point in the bisector of an angle is equidistant from the sides of the angle. 36. The bisectors of the angles contained between the lines 2x — 5y = 0 and 4x + 3y = 12. 37. The... | |
 | George Anthony Hill - Geometry - 1888 - 200 pages
...perpendicular. 4. How are the two theorems on this page related ? 5. By what method are both theorems proved? 6. Theorem. — Every point in the bisector of an...CONCLUSION. P is equally distant from AB and from AC. PROOF. Successive steps : (1) Draw PE JL to AB, PF I. to AC. (2) Prove that A APE — A APF (see p.... | |
 | George Anthony Hill - Geometry - 1889 - 200 pages
...in CD. CONCLUSION. PA = Fit. ADI FIG. 74. PKOOF. Prove that A ADP = A BDP. What follows ? D Fio. 75. B 2. Theorem. — Every point equidistant from the...sides of the angle. HYPOTHESIS. AD the bisector of Z BAG. P any point in AD. CONCLUSION. P is equally distant from AB and from AC. FIQ. 70. PROOF. Successive... | |
 | George Albert Wentworth - 1889 - 264 pages
...ends of a straight line determine the perpendicular through the middle point of the IvaQ.^s" ^£&- Theorem. Every p'oint in the bisector of an angle is equidistant from the sides of the angle, and every point not in the bisector is unequally distant from the sides. 70. Theorem. The line joining... | |
 | George Albert Wentworth - 1889 - 276 pages
...from the ends of a straight line determine the perpendicular through the middle point of the line. 69. Theorem. Every point in the bisector of an angle is equidistant from the sides of the angle, and every point not in the bisector is unequally distant from the sides. 70. Theorem. The line joining... | |
 | George Albert Wentworth - Geometry - 1893 - 270 pages
...the _L). .'. the two A coincide, and are equal. PROPOSITION XXXV. THEOREM. 162. Ever;/ point in tlie bisector of an angle is equidistant from the sides of the angle. Let AD be the bisector of the angle BAC, and let O be any point in AD. To prove that O is equidistant... | |
 | George Albert Wentworth, George Anthony Hill - Geometry - 1894 - 150 pages
...and the hypotenuse of the one are equal respectively to a leg and the hypotenuse of the other. 155. Theorem. Every point in the bisector of an angle is equidistant from the sides of the angle. 156. Theorem. Every point equidistant from the sides of an angle lies in the bisector of the angle.... | |
 | George Albert Wentworth - Mathematics - 1896 - 68 pages
...and the hypotenuse of the one are equal respectively to a side and the hypotenuse of the other. 162. Every point in the bisector of an angle is equidistant from the sides of the angle. 163. Every point within an angle, and equidistant from its sides, is in the bisector of the angle.... | |
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