Search results

  • ...onals are perpendicular, both triangle <math>DMP</math> and triangle <math>BMP</math> are 30-60-90 right triangles.
    13 KB (2,075 words) - 16:50, 9 November 2024
  • ...t, so <math>\angle BOP = \angle BOD/2 = \angle BED</math>, so <math>\angle BMP = \angle BED</math>. Because <math>AC</math> and <math>DE</math> are parall
    4 KB (717 words) - 16:00, 14 April 2024
  • ...rt{3}\right)</math>. At this point, it is useful to note that <math>\Delta BMP</math> is a 30-60-90 triangle with <math>\overline{MB}</math> measuring <ma
    8 KB (1,268 words) - 19:06, 17 October 2024
  • ...cmath><cmath>= \angle BRM - \angle PRM</cmath><cmath>= \angle BRP = \angle BMP.</cmath> ...e PBM - \angle BMP</cmath><cmath>= \left(180^{\circ} - \angle PBM - \angle BMP\right) - \angle MCP</cmath><cmath>= \angle BPM - \angle MCP</cmath><cmath>=
    18 KB (2,485 words) - 21:38, 3 January 2025
  • ...iral similarity, <math> BPM \sim XPY </math> and <math>\angle XYP = \angle BMP = \angle BCA</math>. Hence, <math>\angle XYP = \angle BCA</math>, <math>XY
    3 KB (530 words) - 11:08, 27 March 2022
  • ...endicular from <math>D</math> is <math>N</math>, then right triangle <math>BMP</math> is congruent to right triangle <math>DNP</math>.
    18 KB (2,914 words) - 09:47, 6 January 2025
  • ...since <math>P</math> is the antipode of <math>B</math>, then <math>\angle BMP = 90^{\circ} = \angle EMP = \angle FMP</math> so <math>EP=FP</math> for all
    8 KB (1,485 words) - 21:55, 29 January 2021
  • .... Thus, <math>\triangle AXM \sim \triangle CPM</math>, and <math>\triangle BMP \sim \triangle AMQ</math>.
    4 KB (684 words) - 02:59, 31 August 2024