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  • ...ath> are on <math>\overline{CD}</math> so that <math>DF=1</math> and <math>GC=2</math>. Lines <math>AF</math> and <math>BG</math> intersect at <math>E</m
    13 KB (1,987 words) - 17:53, 10 December 2022
  • ...m]] yields that <math>GC^2 = 12^2 - \sqrt{3}^2 = 141</math>, so <math>EF = GC = \sqrt{141}</math>. Therefore, <math>AB = 5 + 4 + \sqrt{141} = 9 + \sqrt{1
    4 KB (567 words) - 19:20, 3 March 2020
  • ...th>, so <math>\overline{EG} = \frac{4}{5}(15-x)</math> and <math>\overline{GC} = \frac{3}{5}(15-x)</math> so <math>\overline{AG} = \frac{3x+25}{5}</math>
    14 KB (2,340 words) - 15:38, 21 August 2024
  • <cmath>\frac{x}{x+GC}=\frac{14}{17}</cmath> Simplify and we get <math>GC=\frac{3x}{14}</math>.
    9 KB (1,500 words) - 19:06, 8 October 2024
  • <cmath>GE^2 + GC^2 - 2GE \cdot GC \cdot \cos \theta = CE^2 \Longleftrightarrow</cmath><cmath>(2x)^2 + y^2 - (
    11 KB (1,722 words) - 08:49, 13 September 2023
  • ...ath>MGO</math> and <math>AGH'</math>. Since <math>\frac{MG}{GA}=\frac{H'G}{GC} = \frac{1}{2}</math>, and they both share a vertical angle, they are simil Then <math>\frac {DG}{GE} = \frac{1+m}{1+n}, \frac {MG}{GC} = \frac{n+m}{2}.</math>
    59 KB (10,203 words) - 03:47, 30 August 2023
  • ...angle{GAC} = \angle{FAE}</math> because they both correspond to arc <math>{GC}</math>. So <math>\Delta{GBC} \sim \Delta{EAF}</math>.
    6 KB (1,033 words) - 01:36, 19 March 2022
  • ...do the Pythagorean theorem on sides <math>GB, BC, GC</math>. We get <math>GC=\sqrt{5}</math>. We then know that <math>CF=2</math> by Pythagorean theorem <cmath>EGC= \frac{EG\cdot{GC}}{2} = \frac{\sqrt{5x^2 + 5}}{2}</cmath> (the length of CE is calculated wi
    5 KB (811 words) - 14:30, 27 June 2024
  • ...the point at where they meet <math>H</math>. Because <math>\frac{\overline{GC}}{\overline{BE}}</math> = <math>\frac{1}{9}</math>, <math>\frac{\overline{H
    3 KB (520 words) - 18:12, 20 November 2023
  • If <math>\overline {GC} = x</math> then <math>\overline {CD} = 40\sqrt{7}-x</math>, because <math>
    4 KB (648 words) - 16:07, 9 December 2024
  • ...e centroid of <math>\triangle ABC</math> so <math>\vec GA + \vec GB + \vec GC = \vec 0 \implies </math>
    28 KB (4,853 words) - 22:23, 19 November 2024
  • .../math>, and <math>GB=AF=5</math>, so we must have <cmath>GC+5=9\Rightarrow GC=4</cmath> The area of the bottom rectangle is then <cmath>(DC)(GC)=4\times 4=16</cmath>
    4 KB (586 words) - 21:39, 2 October 2024
  • ...ath> are on <math>\overline{CD}</math> so that <math>DF=1</math> and <math>GC=2</math>. Lines <math>AF</math> and <math>BG</math> intersect at <math>E</m
    15 KB (2,167 words) - 18:50, 25 August 2024
  • ...ath> are on <math>\overline{CD}</math> so that <math>DF=1</math> and <math>GC=2</math>. Lines <math>AF</math> and <math>BG</math> intersect at <math>E</m
    5 KB (898 words) - 16:26, 3 July 2023
  • Based on similarity the length of the height of <math>GC</math> is thus <math>\frac{2}{5}\cdot1 = \frac{2}{5}</math>.
    6 KB (937 words) - 10:16, 23 November 2024
  • ...cdot GD}{BD \cdot GC}</math> which is fixed, equation <math>\frac{T(GD)}{T(GC)} = 1,</math> the Claim, and get Point <math>G = BC \cap EF.</math> Prove <math>\frac {BC \cdot GD} {BD \cdot GC}= 2.</math>
    19 KB (3,291 words) - 12:44, 6 October 2024
  • ...until it touches DC if DC was extended both ways by infinity. We know that GC=23. Thus the answer would be the square root of (23 square and (4*square ro
    4 KB (652 words) - 08:18, 23 September 2021
  • ...math>G</math> is the midpoint of <math>\overline{AC}</math> and <math>AG = GC = 5/2</math>. Similarly, we can find from angle chasing that <math>\angle A
    10 KB (1,654 words) - 17:53, 24 November 2024
  • Similarly, <math>GB = \sqrt3</math> and <math>GC = 1</math>. Now, notice that since <math>EC = EF + FG + GC = 1+2+1=4</math>, triangle <math>DEC</math> is equilateral.
    2 KB (308 words) - 18:01, 9 March 2020
  • By pythagorean theorem, BF=<math>4\sqrt{3}</math>, as are DF, EG, and GC.
    3 KB (422 words) - 19:21, 17 December 2024

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