 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 106tekijä(t) Edwin Bidwell Wilson, Josiah Willard Gibbs - 1901 - 436 sivuaKoko teos - Tietoja tästä kirjasta
 | G. S. Pandey - 2002 - 178 sivua
...are perpendicular to each other. Hence the triangle ABC is right-angled at B. Example 5 : Show that the perpendiculars from the vertices of a triangle to the opposite sides are concurrent. Solution : ABC is a triangle and AD and BE are perpendiculars on BC and CA respectively.... | |
 | 480 sivua
...criticise the converses of the following facts: — (i) Congruent triangles have the same area. (ii) The perpendiculars from the vertices of a triangle to the opposite sides are three concurrent lines. 8. In the given figure OX, OY are two roads at right angles and ABCD is... | |
 | University of St. Andrews - 1913 - 1004 sivua
...groups of three, at four points ; and give a proof for one of the groups. Deduce from the figure that the perpendiculars from the vertices of a triangle to the opposite sides are concurrent. 3. If in a triangle the line joining vertex A to any point O cuts the opposite side... | |
| |
