Question

if \\( \overline { c b } \\) bisects \\( \angle a c d \\), what additional information could be used to prove \\( \triangle a b c \cong \triangle d b c \\) using sas? choose three correct answers. \\( c d = 52 \mathrm { cm } \\) \\( \triangle a c d \\) is isosceles with base \\( \overline { a d } \\) \\( m \angle a b c = 125 ^ { \circ } \\) and \\( \overline { a b } \cong \overline { d b } \\) \\( a b = 29 \mathrm { cm } \\) \\( \triangle a b d \\) is isosceles with base \\( \overline { a d } \\)

Explanation:

Step1: Analyze the SAS (Side - Angle - Side) criterion

For \(\triangle ABC\cong\triangle DBC\) using SAS, we need two sides and the included angle to be equal. We know that \(CB\) is common (\(CB = CB\)) and \(\angle ACB=\angle DCB\) (since \(CB\) bisects \(\angle ACD\)).

Step2: Check each option

• Option \(CD = 52\mathrm{cm}\):

If \(CD = 52\mathrm{cm}\), and we know \(AC = 52\mathrm{cm}\) (from the figure), along with \(CB\) common and \(\angle ACB=\angle DCB\), we can use SAS.

• Option \(\triangle ACD\) is isosceles with base \(\overline{AD}\):

If \(\triangle ACD\) is isosceles with base \(AD\), then \(AC = CD\). With \(CB\) common and \(\angle ACB=\angle DCB\), we can use SAS.

• Option \(m\angle ABC = 125^{\circ}\) and \(\overline{AB}\cong\overline{DB}\):

If \(m\angle ABC = 125^{\circ}\), then \(m\angle DBC=125^{\circ}\) (vertically opposite angles). But \(\angle ABC\) and \(\angle DBC\) are not the included angles for SAS.

• Option \(AB = 29\mathrm{cm}\):

If \(AB = 29\mathrm{cm}\), and \(BD = 29\mathrm{cm}\) (from the figure), along with \(CB\) common and \(\angle ACB=\angle DCB\), we can use SAS.

• Option \(\triangle ABD\) is isosceles with base \(\overline{AD}\):

This gives \(AB = BD\), but we still need information about the sides \(AC\) and \(CD\) or the relationship between angles and sides related to the bisected angle for SAS.

Answer:

\(CD = 52\mathrm{cm}\), \(\triangle ACD\) is isosceles with base \(\overline{AD}\), \(AB = 29\mathrm{cm}\)