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the proof that δacb ≅ δecd is shown. given: (overline{ae}) and (overlin…

Question

the proof that δacb ≅ δecd is shown. given: (overline{ae}) and (overline{db}) bisect each other at c. prove: ( \triangle acb cong \triangle ecd ) what is the missing statement in the proof? ( \bigcirc angle bac cong angle dec ) ( \bigcirc angle acd cong angle ecb ) ( \bigcirc angle acb cong angle ecd ) ( \bigcirc angle bca cong angle dca )

Explanation:

Brief Explanations

To prove \(\triangle ACB \cong \triangle ECD\), we use the given that \(\overline{AE}\) and \(\overline{DB}\) bisect each other at \(C\), so \(AC = EC\) and \(BC = DC\) (definition of bisecting segments). For triangle congruence, we need an included angle or another side/angle. The vertical angles formed by the intersection of \(AE\) and \(DB\) at \(C\) are \(\angle ACB\) and \(\angle ECD\). Vertical angles are congruent, so \(\angle ACB \cong \angle ECD\). Let's analyze the options:

  • \(\angle BAC \cong \angle DEC\): These are not necessarily given or deducible from the bisecting info.
  • \(\angle ACD \cong \angle ECB\): These are not the angles needed for the triangles \(\triangle ACB\) and \(\triangle ECD\).
  • \(\angle ACB \cong \angle ECD\): These are vertical angles, so they are congruent, which is the missing statement for SAS (since \(AC = EC\), \(\angle ACB=\angle ECD\), \(BC = DC\)).
  • \(\angle BCA \cong \angle DCA\): These angles are not related to the triangles' congruence here.

Answer:

\(\boldsymbol{\angle ACB \cong \angle ECD}\) (the third option: \(\boldsymbol{\angle ACB \cong \angle ECD}\))