Question

right triangle abc has altitude \\(\overline{cd}\\). move options to the table to complete the proof that the altitude \\(\overline{cd}\\) divides \\(\triangle abc\\) into two similar triangles. \\(\

$$\begin{array}{|c|c|} \\hline \\text{statements} & \\text{reasons} \\\\ \\hline \\overline{cd} \\text{ is the altitude of } \\triangle abc & \\text{given} \\\\ \\hline \\angle bca \\cong \\angle bdc \\cong \\angle cda & \\\\ \\hline \\angle a \\cong \\angle a & \\text{reflexive property} \\\\ \\hline & \\text{angle-angle similarity} \\\\ \\hline \\angle b \\cong \\angle b & \\text{reflexive property} \\\\ \\hline & \\text{angle-angle similarity} \\\\ \\hline & \\text{triangles similar to the same triangle are similar.} \\\\ \\hline \\end{array}$$

\\) corresponding angles are congruent. right angles are congruent. \\(\triangle abc \sim \triangle acd\\) \\(\triangle abc \sim \triangle cbd\\) \\(\triangle acd \sim \triangle cbd\\)

Explanation:

Step1: Analyze ∠BCA ≅ ∠BDC ≅ ∠CDA

Since \( \triangle ABC \) is a right triangle with right angle \( \angle BCA \), and \( \overline{CD} \) is an altitude, \( \angle BDC \) and \( \angle CDA \) are right angles. So the reason for \( \angle BCA \cong \angle BDC \cong \angle CDA \) is "Right angles are congruent."

Step2: Prove \( \triangle ABC \sim \triangle ACD \)

We know \( \angle BCA \cong \angle CDA \) (right angles) and \( \angle A \cong \angle A \) (reflexive property). By Angle - Angle (AA) similarity, \( \triangle ABC \sim \triangle ACD \). So the statement here is \( \triangle ABC \sim \triangle ACD \) with the reason "Angle - angle similarity".

Step3: Prove \( \triangle ABC \sim \triangle CBD \)

We know \( \angle BCA \cong \angle BDC \) (right angles) and \( \angle B \cong \angle B \) (reflexive property). By Angle - Angle (AA) similarity, \( \triangle ABC \sim \triangle CBD \). So the statement here is \( \triangle ABC \sim \triangle CBD \) with the reason "Angle - angle similarity".

Step4: Prove \( \triangle ACD \sim \triangle CBD \)

Since \( \triangle ABC \sim \triangle ACD \) and \( \triangle ABC \sim \triangle CBD \), by the transitive property (triangles similar to the same triangle are similar), we can conclude that \( \triangle ACD \sim \triangle CBD \). So the statement here is \( \triangle ACD \sim \triangle CBD \) with the reason "Triangles similar to the same triangle are similar".

Now, filling the table:

Statements	Reasons
\( \angle BCA \cong \angle BDC \cong \angle CDA \)	Right angles are congruent.
\( \angle A \cong \angle A \)	Reflexive property
\( \triangle ABC \sim \triangle ACD \)	Angle - angle similarity
\( \angle B \cong \angle B \)	Reflexive property
\( \triangle ABC \sim \triangle CBD \)	Angle - angle similarity
\( \triangle ACD \sim \triangle CBD \)	Triangles similar to the same triangle are similar.

Answer:

The completed proof table is as above. The key steps are using the congruence of right angles, reflexive property of angles, AA similarity criterion, and the transitive property of similar triangles. The final similar triangle statement is \( \triangle ACD \sim \triangle CBD \) (along with the other similarity statements for \( \triangle ABC \) with \( \triangle ACD \) and \( \triangle CBD \)).