Question

The side BC of \(\triangle \text{ABC}\) is produced to point D. The bisectors of \(\angle \text{ABC}\) and \(\angle \text{ACD}\) meet at a point E. If \(\angle \text{BAC} = 68^\circ\), then the measure of \(\angle \text{BEC}\) is :

• A. \(35^\circ\)
• B. \(34^\circ\)
• C. \(32^\circ\)
• D. \(30^\circ\)

Hint:

Remember this useful theorem: If one interior angle bisector (like of \(\angle \text{B}\)) and one exterior angle bisector (like of \(\angle \text{ACD}\)) of a triangle meet at a point (E), then the angle formed at that point (\(\angle \text{BEC}\)) is exactly half the third interior angle of the triangle (\(\angle \text{BAC}\)). So, \(\angle \text{BEC} = \frac{1}{2} \angle \text{BAC}\). Given \(\angle \text{BAC} = 68^\circ\), then \(\angle \text{BEC} = 68^\circ / 2 = 34^\circ\).

Answer 1

The Correct Option is B

Concept: There is a standard theorem related to the angle formed by the bisector of an interior angle and the bisector of an exterior angle of a triangle. If the side BC of a \(\triangle \text{ABC}\) is produced to D, and the bisectors of \(\angle \text{ABC}\) (interior angle at B) and \(\angle \text{ACD}\) (exterior angle at C) meet at point E, then: \[ \angle \text{BEC} = \frac{1}{2} \angle \text{BAC} \] Step 1: Identify the given values from the problem and diagram
From the text: \(\angle \text{BAC} = 68^\circ\).
From the diagram and text: BE bisects \(\angle \text{ABC}\).
From the diagram and text: CE bisects the exterior angle \(\angle \text{ACD}\).
These bisectors meet at E. We need to find \(\angle \text{BEC}\). Step 2: Apply the theorem Using the theorem \(\angle \text{BEC} = \frac{1}{2} \angle \text{BAC}\): \[ \angle \text{BEC} = \frac{1}{2} \times 68^\circ \] \[ \angle \text{BEC} = 34^\circ \] Step 3: Compare with the given options The calculated value \(\angle \text{BEC} = 34^\circ\) matches option (2). {Brief justification of the theorem:} In \(\triangle \text{ABC}\), exterior \(\angle \text{ACD} = \angle \text{BAC} + \angle \text{ABC}\). So, \(\angle \text{ECD} = \frac{1}{2} \angle \text{ACD} = \frac{1}{2} (\angle \text{BAC} + \angle \text{ABC})\). In \(\triangle \text{BCE}\), exterior \(\angle \text{ECD} = \angle \text{EBC} + \angle \text{BEC}\). Since \(\angle \text{EBC} = \frac{1}{2} \angle \text{ABC}\), we have \(\frac{1}{2} (\angle \text{BAC} + \angle \text{ABC}) = \frac{1}{2} \angle \text{ABC} + \angle \text{BEC}\). This simplifies to \(\frac{1}{2} \angle \text{BAC} = \angle \text{BEC}\).