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the proof \\( \\triangle a b c \\cong \\triangle d c b \\) that is show…

Question

the proof \\( \triangle a b c \cong \triangle d c b \\) that is shown.
given: \\( \angle a \cong \angle d ; \overline{c d} \\| \overline{a b} \\)
prove: \\( \triangle a b c \cong \triangle d c b \\)
what is the missing reason in the proof?
\

$$\begin{tabular}{cc|c} \\multicolumn{2}{c|}{statement} & reason \\\\ \\hline 1 & \\( \\angle a \\cong \\angle d \\) & 1 given \\\\ 2 & \\( \\overline{c d} \\| \\overline{a b} \\) & 2 given \\\\ 3 & \\( \\overline{c b} \\cong \\overline{b c} \\) & 3 refl prop \\\\ 4 & \\( \\angle a b c \\cong \\angle d c b \\) & 4 alt int. \\( \\angle s \\) are = \\\\ 5 & \\( \\triangle a b c \\cong \\angle d c b \\) & 5? \\\\ \\end{tabular}$$

\\( \bigcirc \\) alt ext. \\( \angle s \\) are =
\\( \bigcirc \\) asa
\\( \bigcirc \\) aas
\\( \bigcirc \\) corr int. \\( \angle s \\) are =

Explanation:

Step1: Recall triangle congruence theorems

  • ASA (Angle - Side - Angle): Two angles and the included side of one triangle are congruent to two angles and the included side of another triangle.
  • AAS (Angle - Angle - Side): Two angles and a non - included side of one triangle are congruent to two angles and the corresponding non - included side of another triangle.

Step2: Analyze the given information

  • We have \(\angle A\cong\angle D\), \(\angle ABC\cong\angle DCB\), and \(CB = BC\) (common side).
  • In this case, the side is not included between the two given angles.

Answer:

AAS