...<cmath>S_{\bigtriangleup{ABM}}+S_{\bigtriangleup{AB'M}}+S_{\bigtriangleup{CDM}}+S_{\bigtriangleup{C'DM}}+S_{\bigtriangleup{B'C'M}}</cmath>
...ector of <math>\angle CDA</math> and <math>DC = DQ</math>, <math>\triangle CDM \cong \triangle QDM</math>.
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14 KB (2,254 words) - 15:46, 6 January 2025
...> so that <math>AD = BD = 15.</math> Given that the area of triangle <math>CDM</math> may be expressed as <math>\frac {m\sqrt {n}}{p},</math> where <math>
...orthogonal projection from <math>C</math> to <math>AB.</math> Thus, <math>[CDM]=\frac{(DM)(MN)} {2}</math>, <math>MN=BN-BM</math>, and <math>[ABC]=\frac{2
...
5 KB (772 words) - 18:47, 1 August 2023
...point of <math> \overline{BD}</math>. What is the area of <math> \triangle CDM</math>?
...}{2}</math> Since <math>\overline{MF}</math> is the height <math>\triangle CDM</math>, the area is <math>\boxed{\textbf{(C) }\frac{\sqrt{3}}{2}}</math>.
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5 KB (904 words) - 14:15, 23 June 2024
...<math>\angle AMD \cong \angle CDM \longrightarrow \angle CMD \cong \angle CDM</math>. Use the [[Base Angle Theorem]] to show <math>\overline{DC} \cong \o
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6 KB (1,003 words) - 19:43, 6 October 2025
...ath>N</math>. <math>d</math> is the [[median]] of triangle <math>\triangle CDM</math>. The formula for the length of a median is <math>m=\sqrt{\frac{2a^2+
Now that we know the sides of <math>\triangle CDM</math>, we proceed to find the length of <math>d</math>.
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2 KB (376 words) - 12:49, 1 August 2022
Let <math>x=\angle CAM</math>, so <math>3x=\angle CDM</math>. Then, <math>\frac{\tan 3x}{\tan x}=\frac{CM/1}{CM/11}=11</math>. Ex
...y the [[Midpoint]] Theorem. So, <math>\angle HBA=x</math> and <math>\angle CDM=\angle CHB=\angle HDA= 3x</math>.
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7 KB (1,179 words) - 19:02, 20 May 2025
...BM = \angle DCM</math> if and only if triangles <math>BCM</math> and <math>CDM</math> are similar, that is
...<math>\angle CMB = \angle DMC</math>, triangles <math>BCM</math> and <math>CDM</math> are similar. Consequently, <math>\frac{CM}{DM} = \frac{BM}{CM}</math
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5 KB (820 words) - 01:39, 10 January 2023
...>\angle AMD = \angle CDM</math>, so <math>\angle AMD = \angle CMD = \angle CDM</math>, so <math>\bigtriangleup CMD</math> is isosceles, and hence <math>CM
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836 bytes (129 words) - 16:39, 20 June 2024
Thus <math>\triangle CNB \sim\triangle CDM</math> where both are <math>15^{\circ}-75^{\circ}-90^{\circ}</math> triangl
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2 KB (378 words) - 08:14, 29 October 2024
...BAC}</math> should be equal to <math>\angle {BDN}</math> and <math>\angle {CDM}</math>. We can quickly prove that the triangles <math>ABC</math>, <math>AP
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13 KB (1,811 words) - 13:36, 1 June 2024
~cdm ~QuickCat (clarified fakesolve element)
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12 KB (1,996 words) - 18:30, 25 October 2025
...e <math>ABCD</math> we construct the equilateral triangles <math>ABK, BCL, CDM, DAN.</math> Prove that the midpoints of the four segments <math>KL, LM, MN
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3 KB (466 words) - 15:54, 29 January 2021
...e <math>ABCD</math> we construct the equilateral triangles <math>ABK, BCL, CDM, DAN.</math> Prove that the midpoints of the four segments <math>KL, LM, MN
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1 KB (199 words) - 15:40, 29 January 2021
...> so that <math>AD = BD = 15.</math> Given that the area of triangle <math>CDM</math> may be expressed as <math>\frac {m\sqrt {n}}{p},</math> where <math>
...
7 KB (1,127 words) - 08:02, 11 July 2023
...point of <math> \overline{BD}</math>. What is the area of <math> \triangle CDM</math>?
...
12 KB (1,874 words) - 09:16, 7 August 2025
<cmath>z+x = \frac {[CDM]}{2[CAM]} = \frac {1}{2(1 + n)}.</cmath>
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59 KB (10,203 words) - 03:47, 30 August 2023
