...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>=
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18 KB (2,485 words) - 21:38, 3 January 2025
...y both intercept the same circular arc <math>PC</math>. Next, <math>\angle BMP = \angle AMC</math> by vertical angles. Hence, we have an Angle-Angle simil
...= 7</math>. We have Angle-Side-Angle congruence that hence triangle <math>BMP</math> is congruent to triangle <math>QMC</math> and hence <math>QM = MP</m
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16 KB (2,559 words) - 14:16, 4 August 2025
...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
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4 KB (735 words) - 13:36, 16 July 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
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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>.
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4 KB (684 words) - 02:59, 31 August 2024
...onals are perpendicular, both triangle <math>DMP</math> and triangle <math>BMP</math> are 30-60-90 right triangles.
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13 KB (2,075 words) - 16:50, 9 November 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
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8 KB (1,268 words) - 19:06, 17 October 2024
...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
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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>.
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22 KB (3,715 words) - 01:08, 26 October 2025
