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Teokset Teokset 31 - 40 / 110 haulle The perpendiculars from the vertices of a triangle to the opposite sides meet in....
" The perpendiculars from the vertices of a triangle to the opposite sides meet in a point. Let the Js be AH, BP, and CK. Through A, B, C suppose B'C', A'C', A'B', drawn II to BC, AC, AB, respectively. Then AH is _L to B'C'. (Why ?) Now ABCB' and A CBC'... "
Vector Analysis: A Text-book for the Use of Students of Mathematics ... - Sivu 106
tekijä(t) Edwin Bidwell Wilson, Josiah Willard Gibbs - 1901 - 436 sivua
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Exercises in Stewart's Plane and Solid Geometry: With Solutions for Teachers

Seth Thayer Stewart - 1893 - 225 sivua
...known, l\ may be constructed, given length of secant — i chord ; whence length of secant is known. i3. The perpendiculars from the vertices of a triangle to the opposite sides are the bisectrices of the angles of the triangle formed by joining the feet of the perpendiculars....

An Examination Manual in Plane Geometry

George Albert Wentworth, George Anthony Hill - 1894 - 138 sivua
...two adjacent angles of one are equal respectively to a side and two adjacent angles of the other. 2. The perpendiculars from the vertices of a triangle to the opposite sides meet in a common point. 3. The interior angle of a regular polygon exceeds the exterior angle by 120°. How many...

Plane and Solid Geometry

Wooster Woodruff Beman, David Eugene Smith - 1895 - 320 sivua
...from c and b, and .'. from a and b. Th. 30 8. .'. P1 lies on CT. Similarly for P4. Th. 30 Theorem 33. The perpendiculars from the vertices of a triangle to the opposite sides are concurrent. Given the A ABC. To prove that the perpendiculars from A, B, C, to a, b, c, respectively,...

Plane and Solid Geometry

Wooster Woodruff Beman, David Eugene Smith - 1895 - 320 sivua
...circumference passes through the feet of the perpendiculars from the other vertices to the opposite sides. 310. The perpendiculars from the vertices of a triangle to the opposite sides bisect the angles of the triangle formed by joining their feet ; the so-called Pedal Triangle. Theorem...

Elements of Geometry: Plane geometry

Andrew Wheeler Phillips, Irving Fisher - 1896
...likewise prove that it is also cut by the third median in the same point. Hence, etc. 145. Exercise. — The perpendiculars from the vertices of a triangle to the opposite sides meet in a point. Hint. — Draw through each vertex a parallel to the opposite side. Prove AE, BH, and CD are perpendicular...

Numerical Problems in Plane Geometry with Metric and Logarithmic Tables

Joe Garner Estill - 1896 - 144 sivua
...triangles are to each other as the squares of their homologous sides. Bowdoin College, June, 1895. 1. The perpendiculars from the vertices of a triangle to the opposite sides meet in a common point. 2. Upon a given straight line describe an arc of a circle which shall contain a given...

Numerical Problems in Plane Geometry: With Metric and Logarithmic Tables

Joe Garner Estill - 1896 - 161 sivua
...triangles are to each other as the squares of their homologous sides. Bowdoin College, June, 1895. 1. The perpendiculars from the vertices of a triangle to the opposite sides meet in a common point. 2. Upon a given straight line describe an arc of a circle which shall contain a given...

Elements of Geometry, Osa 1

Andrew Wheeler Phillips, Irving Fisher - 1896
...likewise prove that it is also cut by the third median in the same point. Hence, etc. 145. Exercise. — The perpendiculars from the vertices of a triangle to the opposite sides meet in a point. flint. — Draw through each vertex a parallel to the opposite side. Prove AE, BH, and CD are perpendicular...

Elements of Geometry: Plane geometry

Andrew Wheeler Phillips, Irving Fisher - 1896
...each parallel to and equal to half of BC. Then prove OE—ON—NC, and DO-OM=MB. 145. Exercise.—The perpendiculars from the vertices of a triangle to the opposite sides meet in a point. Hint.—Draw through each vertex a parallel to the opposite side. Prove AE, BH, and CD are perpendicular...

Plane Geometry

George D. Pettee - 1896 - 253 sivua
...the sides of a triangle meet at a common point. CONCURRENT LINES PROPOSITION XXXV 47 120. Theorem. The perpendiculars from, the vertices of a triangle to the opposite sides 't"eet at a common point. E /G J K Appl. Prove J§ AD, BE, and CF are concurrent Cons. Through vertices...




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