Question

applying theorems about isosceles triangles
cd bisects ∠acb. which statements must be true? choose three
correct answers.
m∠cda = m∠cdb
ac = cd
ad = bd
m∠dca = m∠dac

Explanation:

Step1: Analyze \(m\angle CDA = m\angle CDB\)

Since \(CD\) is the angle - bisector of an isosceles triangle \( \triangle ABC\) (\(AC = BC\)), \(CD\perp AB\) (by the property of isosceles triangle: the angle - bisector, median and altitude from the vertex angle to the base are the same). So, \(\angle CDA=\angle CDB = 90^{\circ}\), and \(m\angle CDA = m\angle CDB\).

Step2: Analyze \(AC = CD\)

There is no information or theorem (such as congruent triangles or special - angle relationships) to suggest that \(AC = CD\). In an isosceles triangle \( \triangle ABC\) with \(AC = BC\), \(CD\) is the angle - bisector (also altitude and median). \(AC\) and \(CD\) are not generally equal.

Step3: Analyze \(AD = BD\)

By the property of an isosceles triangle: the angle - bisector from the vertex angle (\(\angle ACB\)) to the base (\(AB\)) is also the median. So, \(AD = BD\).

Step4: Analyze \(m\angle DCA=m\angle DAC\)

\(m\angle DCA\) is half of \(\angle ACB\) (because \(CD\) is the angle - bisector), and \(m\angle DAC=\frac{1}{2}(180^{\circ}-\angle ACB)\) (using the angle - sum property of a triangle \(\angle A+\angle B+\angle C = 180^{\circ}\) and \(\angle A=\angle B\) in \(\triangle ABC\)). So, \(m\angle DCA
eq m\angle DAC\).

Answer:

\(m\angle CDA = m\angle CDB\), \(AD = BD\)