Question

which piece of additional information can be used to prove \\( \triangle cea - \triangle cdb \\)?
\\( \angle bdc \\) and \\( \angle aed \\) are right angles
\\( \overline { ae } = \overline { ed } \\)
\\( \triangle bdc \\) is a right triangle
\\( \angle dbc = \angle dcb \\)

Explanation:

Step1: Recall the AA (Angle - Angle) similarity criterion

For two triangles \(\triangle CEA\) and \(\triangle CDB\), we already have a common angle \(\angle C\) (i.e., \(\angle ECA=\angle DCB\)). If we can show that another pair of angles are equal, we can use the AA similarity criterion.

Step2: Analyze each option

• Option 1: If \(\angle BDC\) and \(\angle AED\) are right angles. Then \(\angle BDC=\angle AED = 90^{\circ}\). Along with the common angle \(\angle C\), by AA (Angle - Angle) similarity criterion \(\triangle CEA\sim\triangle CDB\).
• Option 2: \(\overline{AE}\cong\overline{ED}\) gives information about the sides of \(\triangle AED\), not directly related to the similarity of \(\triangle CEA\) and \(\triangle CDB\).
• Option 3: Saying \(\triangle BDC\) is a right - triangle only tells us one triangle is right - angled. We need information about angles of both \(\triangle CEA\) and \(\triangle CDB\) for similarity.
• Option 4: \(\angle DBC=\angle DCB\) gives information about the angles of \(\triangle BDC\) (isosceles triangle property), but we need an angle relation between \(\triangle CEA\) and \(\triangle CDB\) (other than \(\angle C\)) for similarity.

Answer:

\(\angle BDC\) and \(\angle AED\) are right angles.