Question

which piece of additional information can be used to prove △cea ~ △cdb?
○ ∠bdc and ∠aed are right angles
○ \\(\overline{ae} \cong \overline{ed}\\)
○ △bdc is a right triangle
○ ∠dbc ≅ ∠dcb

Explanation:

Step1: Recall Similar Triangle Criteria

To prove two triangles similar, we can use AA (Angle - Angle) criterion, SAS (Side - Angle - Side) criterion, or SSS (Side - Side - Side) criterion. For \(\triangle CEA\) and \(\triangle CDB\), we already know that \(\angle C\) is common to both triangles. So we need another pair of congruent angles.

Step2: Analyze Each Option

• Option 1: \(\angle BDC\) and \(\angle AED\) are right angles. If \(\angle BDC = 90^{\circ}\) and \(\angle AED=90^{\circ}\), then \(\angle AEC = 90^{\circ}\) (since \(\angle AED\) is a straight - line related angle? Wait, actually, \(\angle AEC\) and \(\angle BDC\) would both be right angles. Since \(\angle C\) is common, by AA (Angle - Angle) criterion, if \(\angle AEC=\angle BDC = 90^{\circ}\) and \(\angle C=\angle C\), then \(\triangle CEA\sim\triangle CDB\).
• Option 2: \(\overline{AE}\cong\overline{ED}\). This gives information about segments in the left - hand part of the figure and does not directly relate to the angles or sides needed to prove similarity between \(\triangle CEA\) and \(\triangle CDB\).
• Option 3: \(\triangle BDC\) is a right triangle. Just knowing \(\triangle BDC\) is a right triangle does not tell us that \(\triangle CEA\) is also a right triangle (we need the corresponding angles to be equal for AA).
• Option 4: \(\angle DBC\cong\angle DCB\). This tells us that \(\triangle BDC\) is isosceles, but it does not give us a corresponding angle equal in \(\triangle CEA\) to use for the AA criterion.

Answer:

\(\boldsymbol{\angle BDC}\) and \(\boldsymbol{\angle AED}\) are right angles (the first option)