(Proof)
\(\fbox{\(\;(\alpha)\;\)}\) from triangle \(\mathrm{ACD}\)
\((1)\)
\(\fbox{\(\;(\beta)\;\)}\) from triangle \(\mathrm{BCE}\)
\((2)\)
Since \(\angle \mathrm{ACD} = \angle \mathrm{ECB} = 60°\),
\(\angle \mathrm{ACE} = 60°+ \angle \mathrm{DCE} = \angle \mathrm{DCB}\)
\((3)\)
From \((1), (2)\) and \((3)\),
\(\triangle \mathrm{ACE} \equiv \triangle \mathrm{DCB}\)
since the two sides and the angle between them are equivalent.
Therefore \(\overline{\mathrm{AE}} = \overline{\mathrm{DB}}\).
In the proof above, what are appropriate for \((\alpha)\) and \((\beta)\)?
|
\((\alpha)\) |
\((\beta)\) |
① |
\(\overline{\mathrm{AC}} = \overline{\mathrm{AD}}\) |
\(\overline{\mathrm{CE}} = \overline{\mathrm{BE}}\) |
② |
\(\overline{\mathrm{AC}} = \overline{\mathrm{DC}}\) |
\(\overline{\mathrm{CE}} = \overline{\mathrm{BE}}\) |
③ |
\(\overline{\mathrm{AD}} = \overline{\mathrm{CD}}\) |
\(\overline{\mathrm{CB}} = \overline{\mathrm{BE}}\) |
④ |
\(\overline{\mathrm{AC}} = \overline{\mathrm{AD}}\) |
\(\overline{\mathrm{CE}} = \overline{\mathrm{CB}}\) |
⑤ |
\(\overline{\mathrm{AC}} = \overline{\mathrm{DC}}\) |
\(\overline{\mathrm{CE}} = \overline{\mathrm{CB}}\) |