Question

if cb bisects ∠acd, what additional information could be used to prove △abc ≅ △dbc using sas? select three options. m∠abc = 125° and ab ≅ db △acd is isosceles with base ad △abd is isosceles with base ad cd = 52 cm ab = 29 cm

Explanation:

Step1: Analyze the first option

If \(m\angle ABC = 125^{\circ}\) and \(\overline{AB}\cong\overline{DB}\), we know that \(\angle ABC+\angle DBC = 180^{\circ}\) (linear - pair), so \(m\angle DBC=55^{\circ}
eq125^{\circ}\). This option is incorrect for SAS.

Step2: Analyze the second option

If \(\triangle ACD\) is isosceles with base \(\overline{AD}\), then \(AC = CD\). Since \(\overline{CB}\) bisects \(\angle ACD\), \(\angle ACB=\angle DCB\). And \(CB = CB\) (common side). By SAS (\(AC = CD\), \(\angle ACB=\angle DCB\), \(CB = CB\)), \(\triangle ABC\cong\triangle DBC\).

Step3: Analyze the third option

If \(\triangle ABD\) is isosceles with base \(\overline{AD}\), then \(AB = DB\). Since \(\overline{CB}\) bisects \(\angle ACD\), \(\angle ACB=\angle DCB\). And \(CB = CB\) (common side). By SAS (\(AB = DB\), \(\angle ACB=\angle DCB\), \(CB = CB\)), \(\triangle ABC\cong\triangle DBC\).

Step4: Analyze the fourth option

If \(CD = 52\mathrm{cm}\), then \(AC = CD\) (given \(AC = 52\mathrm{cm}\)). Since \(\overline{CB}\) bisects \(\angle ACD\), \(\angle ACB=\angle DCB\). And \(CB = CB\) (common side). By SAS (\(AC = CD\), \(\angle ACB=\angle DCB\), \(CB = CB\)), \(\triangle ABC\cong\triangle DBC\).

Step5: Analyze the fifth option

If \(AB = 29\mathrm{cm}\), we know \(BD = 29\mathrm{cm}\) (given \(BD = 29\mathrm{cm}\)). But we don't know the relationship between the angles involving \(AB\) and \(BD\) in the context of the angle - bisected \(\angle ACD\) for the SAS criterion. This option is incorrect.

Answer:

\(\triangle ACD\) is isosceles with base \(\overline{AD}\), \(\triangle ABD\) is isosceles with base \(\overline{AD}\), \(CD = 52\mathrm{cm}\)