Question

△abc and △cde are shown below.
which statement is true?
△abc is similar to △cde.
△abc is not similar to △cde.
there is not enough information to determine whether the

Explanation:

Step1: Find angle at C in ΔABC

In ΔABC, sum of angles is \(180^\circ\). So \(\angle ACB = 180^\circ - 62^\circ - 41^\circ = 77^\circ\).

Step2: Check angles in ΔCDE

In ΔCDE, one angle is \(84^\circ\), and let's check if angles match ΔABC. The angles of ΔABC are \(62^\circ\), \(41^\circ\), \(77^\circ\); angles of ΔCDE (we can find the third angle, but even with the given \(84^\circ\) and the marked parallel - like arrows (maybe alternate interior, but let's check angle correspondence). Wait, actually, the key is angle - angle similarity. Let's re - calculate ΔABC's angles: \(180-(62 + 41)=77\). Now, in ΔCDE, if we assume the arrowed sides are parallel, but actually, let's check the angles. Wait, maybe I made a mistake. Wait, no: in ΔABC, angles are \(62^\circ\) (at A), \(41^\circ\) (at B), so at C: \(180 - 62 - 41=77^\circ\). In ΔCDE, angle at E is \(84^\circ\). Now, let's see the other angles. Wait, maybe the arrowed sides imply that \(AB\parallel CE\)? No, the arrows are on \(AB\) and \(CE\)? Wait, the diagram has arrows on \(AB\) (from A to B? No, the arrows are on the sides: one on \(AB\) (from A to B? Wait, no, the arrows are on \(AB\) (the side from A to B) and on \(CE\) (the side from C to E). So maybe \(AB\parallel CE\), so \(\angle BAC=\angle ECD\)? Wait, \(\angle BAC = 62^\circ\), so if \(AB\parallel CE\), then \(\angle ECD=\angle BAC = 62^\circ\). Then in ΔCDE, angle at C is \(62^\circ\), angle at E is \(84^\circ\), so angle at D is \(180 - 62 - 84 = 34^\circ\). But in ΔABC, angles are \(62^\circ\), \(41^\circ\), \(77^\circ\). So the angles don't match. Wait, no, maybe I messed up the angle at C in ΔABC. Wait, no: \(62+41 = 103\), \(180 - 103 = 77\). Now, in ΔCDE, if the arrowed sides are \(AB\) and \(CE\), so corresponding angles: \(\angle BAC=\angle ECD = 62^\circ\) (alternate interior angles). Then in ΔCDE, angle at C is \(62^\circ\), angle at E is \(84^\circ\), so angle at D is \(180-(62 + 84)=34^\circ\). Now, in ΔABC, angles are \(62^\circ\), \(41^\circ\), \(77^\circ\); in ΔCDE, angles are \(62^\circ\), \(84^\circ\), \(34^\circ\). No two angles are equal between the triangles, so they are not similar. Wait, but maybe I made a mistake. Wait, let's re - check. Wait, the problem is about similarity. For two triangles to be similar, two angles must be equal (AA similarity). In ΔABC: angles are \(62^\circ\) (A), \(41^\circ\) (B), \(77^\circ\) (C). In ΔCDE: let's find its angles. If we consider the straight line A - C - D, so \(\angle ACB+\angle ECD+\angle\) (wait, no, A - C - D is a straight line, so \(\angle ACB+\angle ECD+\) no, \(\angle ACB\) and \(\angle ECD\) are adjacent? Wait, no, A, C, D are colinear, so \(\angle ACB\) and \(\angle ECD\) are supplementary? No, A - C - D is a straight line, so \(\angle ACB+\angle ECD+\) no, \(\angle ACB\) is at C between A and B, \(\angle ECD\) is at C between E and D. So they are adjacent angles on the straight line, so \(\angle ACB+\angle ECD+\angle BCE = 180^\circ\)? No, no, A - C - D is a straight line, so \(\angle ACB+\angle ECD+\) no, \(\angle ACB\) and \(\angle ECD\) are on the line A - C - D, with B and E on different sides. So actually, \(\angle ACB+\angle ECD+\) no, it's \(\angle ACB+\angle BCE+\angle ECD = 180^\circ\)? No, I think I misread the diagram. Let's start over.

In triangle ABC: angles are \(\angle A = 62^\circ\), \(\angle B = 41^\circ\), so \(\angle C=180 - 62 - 41 = 77^\circ\).

In triangle CDE: angle at E is \(84^\circ\). Now, if we assume that the sides with arrows (AB and CE) are parallel, then \(\angle A=\angle ECD\) (alternate interior angles), so \(\angle ECD = 62^\circ\). Then in triangle CDE, \(\angle D=180 - 62 - 84 = 34^\circ\).

Now, the angles of triangle ABC are \(62^\circ\), \(41^\circ\), \(77^\circ\); angles of triangle CDE are \(62^\circ\), \(84^\circ\), \(34^\circ\). There is no pair of two equal angles between the two triangles. So by AA similarity criterion, since no two angles are equal, the triangles are not similar.

Answer:

\(\triangle ABC\) is not similar to \(\triangle CDE\).