Question

ray ce is the angle bisector of ∠acd. which statement about the figure must be true?
○ $m\angle ecd = \frac{1}{2}m\angle ecb$
○ $m\angle ace = \frac{1}{2}m\angle acd$
○ $\angle ace \cong \angle dcb$
○ $\angle ecd \cong \angle acd$

Explanation:

An angle bisector divides an angle into two equal - measure angles. Since ray \(CE\) is the angle bisector of \(\angle ACD\), by the definition of an angle bisector, \(\angle ACE\) and \(\angle ECD\) are equal, and each of them is equal to \(\frac{1}{2}\) of \(\angle ACD\).

• For the first option: There is no relationship given that would imply \(m\angle ECD=\frac{1}{2}m\angle ECB\). We don't know the measure of \(\angle ECB\) relative to \(\angle ECD\) from the angle - bisector information.
• For the second option: Since \(CE\) bisects \(\angle ACD\), \(m\angle ACE = m\angle ECD=\frac{1}{2}m\angle ACD\), so this statement is true.
• For the third option: There is no reason to believe that \(\angle ACE\cong\angle DCB\). The angle - bisector only gives information about the angles formed by \(CE\) with \(AC\) and \(CD\), not about the relationship between \(\angle ACE\) and \(\angle DCB\).
• For the fourth option: \(\angle ECD\) is a part of \(\angle ACD\) (since \(CE\) is inside \(\angle ACD\) as it is the bisector), so \(m\angle ECD=\frac{1}{2}m\angle ACD\), and \(\angle ECD\) cannot be congruent to \(\angle ACD\) (unless \(\angle ACD = 0^{\circ}\), which is not the case here).

Answer:

\(m\angle ACE=\frac{1}{2}m\angle ACD\) (the second option)