...sides of a triangle whose base has been proved to be E — 2r. Its length is therefore ^R — r. (41 Hence, when the inscribed and circumscribing circles...belongs to M. Terquem, who has given it in his Annales, t. I. p. 196. Dr. Hart has since made other more direct demonstrations of it, and Sir William R. Hamilton,...

...of the sides of a triangle whose base has been proved to be R — 2r. Its length is therefore ^B — r. (4) Hence, when the inscribed and circumscribing...belongs to M. Terquem, who has given it in his Annales, t. I. p. 196. Dr. Hart has since made other more direct demonstrations of it, and Sir William R. Hamilton,...

...object of this Paper is to give a simple geometrical proof of the following theorem : " The circle through the middle points of the sides of a triangle touches the inscribed circle and the three escribed circles." This theorem has not, I believe, been hitherto proved...

...and the circle with respect to which that triangle is self-conjugate. 306. To find the equation to the circle which passes through the middle points of the sides of the triangle of reference. Let the equation be (Art. 293) a" sin 2A + ¿S" sin 2B + y* sin 2 G = (la...

...and d' are the diagonals, and H is the radius of the circumscribing circle. 12. Find the equation of the circle which passes through the middle points of the sides of the triangle a = o, /3 = o, y=o; and prove that it touches the circle inscribed in the given triangle....

...a/3y + frya+ca/9 = 5 (aa + 5/3 + 07) (a cos^ +/9 cos 5 + 7 cos (7). 2 347. To find the equation to the circle, which passes through the middle points of the sides of the triangle ABO. Let the equation be 01/87 + 67a + ca/3 = (aa. + 5/3 + 07) (?a + m/3 + ny) ; then,...

...mß + ny), provided we properly determine the constants in each case. 329. To express the equation to the circle which passes through the middle points of the sides of the triangle of reference. Let a = 0, /3 = 0, 7 = 0 be the equations to the straight lines which form...

...= —«. The resolution of these equations gives M = 2a cos A, v = 26 cos B, w — 2c cos C. Hence the circle which passes through the middle points of the sides of reference {the nine-point circle) becomes a2 sin A cos A + /32 sin B cos B + y2 sin (7 cos C — /3y...

...radius of the circumscribing circle. QED NB For a geometrical proof of the theorem that " the circle through the middle points of the sides of a triangle touches the inscribed circle and the three escribed circles" the reader is referred to a paper by the author of...

...forcible illustration of this conjunction will be found in the "six points circle," as it has been called, which passes through the middle points of the sides of a triangle, passes also through the feet of the three perpendiculars let fall from the angles upon the opposite...