Question

cd bisects ∠acb. which statements must be true? check all that apply. ad = bd ac = cd m∠acd = m∠bcd m∠cda = m∠cdb m∠dca = m∠dac

Explanation:

1. For \(AD = BD\): The triangle has \(AC = BC\) (marked congruent), but \(CD\) is an angle bisector, not necessarily a median unless the triangle is isosceles in a way that \(CD\) is also a median. However, just from angle bisector, we can't conclude \(AD = BD\) without more info.
2. For \(AC = CD\): There's no given or implied reason for \(AC\) and \(CD\) to be equal.
3. For \(m\angle ACD=m\angle BCD\): By definition of an angle bisector, \(CD\) bisects \(\angle ACB\), so it splits it into two equal angles. Thus, \(m\angle ACD = m\angle BCD\) must be true.
4. For \(m\angle CDA=m\angle CDB\): Since \(AC = BC\), \(\triangle ACB\) is isosceles with \(AC = BC\), and \(CD\) is the angle bisector (and in an isosceles triangle, the angle bisector from the apex is also the median and altitude). So \(\angle CDA\) and \(\angle CDB\) are right angles (if we consider the altitude property) or at least equal because \(\triangle CDA\cong\triangle CDB\) (by SAS: \(AC = BC\), \(\angle ACD=\angle BCD\), \(CD = CD\)). So \(m\angle CDA=m\angle CDB\) is true.
5. For \(m\angle DCA=m\angle DAC\): There's no given or implied reason for these two angles to be equal.

Answer:

• \(m\angle ACD = m\angle BCD\)
• \(m\angle CDA = m\angle CDB\)