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  • ...plane <math>A'B'C'D'</math> intersect the plane at <math>O''</math>. By HL congruency, the triangles <math>OO''A'</math>, <math>OO''B'</math>, <math>OO''C'</math
    3 KB (509 words) - 22:22, 15 August 2012
  • '''Congruency''' is a property of multiple geometric figures. ...ngle AB'C</math>. We have <math>\triangle ABC\cong\triangle AB'C</math> by SAS congruence.
    10 KB (1,655 words) - 15:56, 17 September 2024
  • ...BC}</math>~<math>\triangle{CDE}</math>~<math>\triangle{EFA}</math> via SAS congruency. Using the formula <math>[ABC]=\frac{ab \sin C}{2}= \frac{r \sqrt{3}}{4}</m ...th>\triangle{CDE}</math>, and <math>\triangle{EFA}</math> are congruent by SAS, so the side opposite the 120 degree angle is also the same (since the tria
    4 KB (702 words) - 19:17, 9 October 2024
  • ...h>\triangle BFS</math> and <math>\triangle BFA</math> are congruent by ASA congruency, and so are <math>\triangle CGT</math> and <math>\triangle CGA</math>. We h ...>c</math> and one angle of measure <math>90^\circ - \dfrac{B}{2}</math> by SAS Congruence, and so <math>\angle{BFA} = 90^\circ</math>. It then follows tha
    7 KB (1,189 words) - 00:22, 19 November 2023
  • ...mma</math>, <math>\alpha</math>, and <math>\beta</math>. It follows by SAS congruency that this triangle is congruent to <math>FAB</math>, <math>BCD</math>, and
    11 KB (1,925 words) - 11:07, 31 August 2023
  • ...<math>\angle BEO = \angle CDO</math>. Since <math>OD = OE</math>, by SAS Congruency, <math>\triangle BEO \cong \triangle CDO</math>, so <math>OB = OC</math>. By SSS Congruency, <math>\triangle ABO \cong \triangle ACO</math>, so <math>\angle BAO = \ang
    3 KB (539 words) - 20:01, 29 July 2018
  • ...h>D</math> be on <math>FE</math> so that <math>FD = DB</math>, then by SAS Congruency, <math>\triangle DMB \cong \triangle DMC</math>, so <math>FD = DB = DC</mat
    2 KB (410 words) - 20:18, 31 May 2020
  • ...C = AB</math>. Thus, <math>\triangle CAD \cong \triangle BAD</math> by SAS Congruency, so <math>AD \perp BC</math> and <math>BD = CD</math>.
    4 KB (610 words) - 23:43, 13 January 2019
  • Draw lines <math>AC</math> and <math>BD</math>. By SAS Similarity, <math>\triangle DSR \sim \triangle DAC,</math> <math>\triangle ...\angle RPQ</math> and <math>\angle RSQ = \angle SQP.</math> Thus, by ASA Congruency, <math>\triangle SRO \cong \triangle QPO.</math> Finally, using CPCTC show
    3 KB (583 words) - 16:52, 22 January 2024
  • ...ath>\angle ACB,</math> <math>\angle ACX = \angle BCX.</math> Thus, by AAS Congruency, <math>\triangle ACX \cong \triangle BCX,</math> which results in <math>AC ...ngle B'CY,</math> so <math>\triangle ACY \cong \triangle BCY</math> by AAS Congruency, making <math>AC = BC.</math> However, it is given that <math>AC \ne BC,</
    3 KB (394 words) - 23:27, 13 August 2018
  • ...NO = \angle NOR,</math> making <math>\triangle MOS \cong NOR</math> by SAS Congruency. That means <math>SM = NR = s</math> by CPCTC. .../math> be the intersection of <math>MN</math> and <math>PR.</math> By SSS Congruency, <math>\triangle POQ \cong \triangle QON \cong \triangle NOR,</math> so <ma
    7 KB (1,181 words) - 19:41, 17 April 2022
  • Since <math>AE = EC</math> and <math>BD = DC</math>, by SAS Similarity, <math>\triangle EDC \sim \triangle ACB</math>. From the simila ...D</math> is the midpoint of <math>CB</math>, <math>CD = DB</math>. By SAS Congruency, <math>\triangle ACD \cong \triangle ABD</math>, so <math>\angle CAD = \ang
    2 KB (365 words) - 11:55, 4 December 2018
  • ...>N</math> to <math>P</math> and <math>Q</math>, as seen in the diagram. By SAS Similarity, we find that <math>\triangle CBA \sim \triangle CQM</math> and ...<math>S</math> be on <math>AD</math> where <math>IS \perp AD</math>. By HL Congruency, <math>\triangle ITQ \cong \triangle ISP</math>, so <math>PS = QT</math>. N
    3 KB (586 words) - 22:47, 8 January 2019
  • Let <math>E</math> be the midpoint of <math>CD</math>. By SSS Congruency, <math>\triangle OCE \cong \triangle ODE</math>. Thus, <math>\angle OEC = ...<math>AB</math>, and let <math>F</math> be the point of intersection. By SAS Similarity, <math>\triangle AOB \sim \triangle COD</math>, so <math>\angle
    4 KB (710 words) - 13:01, 23 December 2019
  • Therefore <math>A[\Delta AED] \cong A[\Delta BEG]</math> (SAS Congruency Theorem).
    27 KB (4,265 words) - 00:25, 9 December 2024
  • * [[SAS Similarity]] * [[SAS Congruency]]
    795 bytes (113 words) - 13:02, 29 December 2023
  • ...math>EO=EO</math>, and <math>\angle QEO=\angle PEO=90^\circ</math>. By SAS congruency, <math>\Delta QOE\cong\Delta POE\implies QO=PO</math>. Similarly, we find t
    12 KB (1,874 words) - 14:42, 30 April 2024