Proceedings of the Edinburgh Mathematical Society, Volúmenes1-2Scottish Academic Press, 1894 |
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Página 14
... similar . Also , since the lines AA ' , BB ' , CC ' joining corresponding vertices are concurrent at G , triangles ABC , A'B'C ' are homothetic , and G is their homothetic centre . DEF . - Triangles such as the fundamental triangle ABC ...
... similar . Also , since the lines AA ' , BB ' , CC ' joining corresponding vertices are concurrent at G , triangles ABC , A'B'C ' are homothetic , and G is their homothetic centre . DEF . - Triangles such as the fundamental triangle ABC ...
Página 15
... similar to triangle ABC . ( 10 ) If P be any point on the circumcircle of ABC , the centroids of the four triangles PBC , PCA , PAB , ABC are concyclic . ‡ For if the centroids of these triangles be denoted by D , E , F , G respectively ...
... similar to triangle ABC . ( 10 ) If P be any point on the circumcircle of ABC , the centroids of the four triangles PBC , PCA , PAB , ABC are concyclic . ‡ For if the centroids of these triangles be denoted by D , E , F , G respectively ...
Página 18
... similar ; therefore therefore BB ' : NM ' = - BC ' : NC ' = BC : MC = m + n : m ; mBB ' = ( m + n ) NM ' . The triangles BCC ' , BMN are similar ; = CC ' : MN- BC : BM , n ; therefore therefore Hence = m + n : nCC ' = ( m + n ) MN . mBB ...
... similar ; therefore therefore BB ' : NM ' = - BC ' : NC ' = BC : MC = m + n : m ; mBB ' = ( m + n ) NM ' . The triangles BCC ' , BMN are similar ; = CC ' : MN- BC : BM , n ; therefore therefore Hence = m + n : nCC ' = ( m + n ) MN . mBB ...
Página 20
... similar and oppositely situated , G is their homothetic centre , and 2 : 1 is the ratio of similitude . Hence if AD , BE , CF be concurrent at O , the corresponding straight lines A'D ' , B'E ' , C'F ' will pass through the ...
... similar and oppositely situated , G is their homothetic centre , and 2 : 1 is the ratio of similitude . Hence if AD , BE , CF be concurrent at O , the corresponding straight lines A'D ' , B'E ' , C'F ' will pass through the ...
Página 26
... similar to each other ... and DEF , D1E1F1 99 99 " " " ; 99 99 FIGURE 14 . * The proof of the theorem will appear from the figure † if it be observed that correspond to ( 4 , 5 ; 6 , 1 ; 2 , 3 ) , ( 1 , 2 , 3 , 4 ; 5 , 6 ) , ; Triangles ...
... similar to each other ... and DEF , D1E1F1 99 99 " " " ; 99 99 FIGURE 14 . * The proof of the theorem will appear from the figure † if it be observed that correspond to ( 4 , 5 ; 6 , 1 ; 2 , 3 ) , ( 1 , 2 , 3 , 4 ; 5 , 6 ) , ; Triangles ...
Otras ediciones - Ver todas
Proceedings of the Edinburgh Mathematical Society Edinburgh Mathematical Society Vista completa - 1966 |
Términos y frases comunes
2abcA a+b+c A₁ B₁ bisects centroid circle circumcentre circumcircle circumscribed collinear complementary triangle concyclic congruent conic contour-lines curve D₁ denoted diagonal diameter double point draw drawn Edinburgh Mathematical Society enantiomorph equal excentres excircle FIGURE formed by joining four triangles fundamental triangle Gentleman's Diary Geometry George Watson's College given H₁ Hence incentre incircle inscribed internal bisector intersect Lady's and Gentleman's m₁ Mathematical Master Mathématiques medians meet the circumcircle mid point nine-point circle oppositely situated orthic triangle orthocentre pairs parallel to BC parallelogram passes perpendicular to BC plane point of BC polar Professor proof properties Proposition quadrilateral r₁ radii radius respectively rhombi right angles semiperimeter similar Similarly spherical tangents tartaric acid temperature theorem THOMAS MUIR triangle ABC triangle formed vertex vertices
Pasajes populares
Página 52 - The diagonals of a quadrilateral intersect at right angles. Prove that the sum of the squares on one pair of opposite sides is equal to the sum of the squares on the other pair.
Página 11 - Any two sides of a triangle are together greater than the third side.
Página 47 - The six straight lines. joining two and two the centres of the four circles which touch the sides of a triangle pass each through one of the vertices of the triangle.
Página 15 - The perpendiculars from the vertices of a triangle on the opposite sides are concurrent.
Página 34 - Compare the area of the triangle formed by joining the centres of these squares with the area of the equilateral triangle.
Página 11 - The opposite angles of a quadrilateral inscribed in a circle are together equal to two right angles, with converse.
Página 37 - If two triangles have two sides of the one equal to two sides of the...
Página 14 - Wherefore, if the ratios &c. EXERCISES. 1. Shew that the locus of the middle points of straight lines parallel to the base of a triangle and terminated by its sides is a straight line. 2. CAB, CEB are two triangles having the angle B common and the sides CA, CE equal ; if BAE be produced to D and ED be taken a third proportional to BA , AC, then the triangle BDC is similar to the triangle BAC.
Página 8 - The straight line, joining the points of bisection of two sides of a triangle, is parallel to the base.
Página 62 - If the perpendicular from any vertex of a triangle to the opposite side divides that side into two segments, how does each of these segments compare in length with its adjacent side of the triangle ? Prove it.