In triangle ABC shown in the figure, side BC is extended to point E, and CD bisects the external angle...

1. TRANSLATE the problem information

Given:

• Triangle ABC with \(\angle\mathrm{BAC} = 65°\) and \(\angle\mathrm{ABC} = 38°\)
• Side BC is extended beyond C to point E (creating an exterior angle)
• Ray CD bisects the exterior angle \(\angle\mathrm{ACE}\)
• Find: \(\angle\mathrm{DCE}\) (half of the exterior angle)

What we need to find: The measure of \(\angle\mathrm{DCE}\), which is half of the exterior angle \(\angle\mathrm{ACE}\).

2. INFER the solution strategy

Since we need half of an angle (\(\angle\mathrm{DCE}\) is half of \(\angle\mathrm{ACE}\) because CD bisects it), we must first find the full exterior angle \(\angle\mathrm{ACE}\).

Key strategic insight: We can find the exterior angle using the Exterior Angle Theorem without first finding all interior angles - though finding \(\angle\mathrm{ACB}\) first also works.

3. Find the interior angle \(\angle\mathrm{ACB}\) (optional but helpful)

Using the angle sum property of triangles:

\(\angle\mathrm{ACB} = 180° - \angle\mathrm{BAC} - \angle\mathrm{ABC}\)
\(\angle\mathrm{ACB} = 180° - 65° - 38° = 77°\)

4. INFER and apply the Exterior Angle Theorem

The exterior angle at any vertex equals the sum of the two non-adjacent interior angles.

Therefore:

\(\angle\mathrm{ACE} = \angle\mathrm{BAC} + \angle\mathrm{ABC}\)
\(\angle\mathrm{ACE} = 65° + 38° = 103°\)

(Verification: \(\angle\mathrm{ACE}\) should also equal \(180° - \angle\mathrm{ACB} = 180° - 77° = 103°\) ✓)

5. Apply the angle bisector definition

Since CD bisects \(\angle\mathrm{ACE}\), it divides the angle into two equal parts:

\(\angle\mathrm{DCE} = \angle\mathrm{ACE} ÷ 2\)
\(\angle\mathrm{DCE} = 103° ÷ 2 = 51.5°\)

Answer: 51.5° or 51.5 or 103/2

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill combined with conceptual confusion: Students misunderstand which angle is being bisected. They might bisect the interior angle \(\angle\mathrm{ACB}\) instead of the exterior angle \(\angle\mathrm{ACE}\).

This leads to calculating: \(77° ÷ 2 = 38.5°\), which is incorrect. The problem specifically states "CD bisects the external angle \(\angle\mathrm{ACE}\)," not the interior angle at C. This confusion stems from not carefully visualizing or understanding the geometric setup where BC is extended to create an exterior angle.

Second Most Common Error:

Missing conceptual knowledge or weak INFER skill: Students don't recall or recognize the Exterior Angle Theorem and get stuck trying to find \(\angle\mathrm{ACE}\). They might attempt to use supplementary angles (\(180° - 77° = 103°\)) which works, but without understanding why that exterior angle exists, they may make errors in the setup or lose confidence in their answer.

Some students might also forget that the exterior angle exists at all and try to work only with the three interior angles, never arriving at the correct angle to bisect.

The Bottom Line:

This problem tests whether students can navigate a multi-step geometry problem involving both interior and exterior angles. Success requires careful translation of the geometric setup (understanding what "extended" and "bisects the external angle" mean) and strategic application of the Exterior Angle Theorem. The decimal answer (51.5°) also confirms that bisecting is the final step, not finding a whole-number angle measure.