Question

given: $overline{ab}congoverline{ac}$ and $angle badcongangle cad$. prove: $angle dbccongangle dcb$. note: quadrilateral properties are not permitted in this proof. step statement reason 1 $overline{ab}congoverline{ac}$ $angle badcongangle cad$ given

Explanation:

Step1: Prove $\triangle ABD\cong\triangle ACD$

Since $\overline{AB}\cong\overline{AC}$, $\angle BAD\cong\angle CAD$, and $\overline{AD}=\overline{AD}$ (common - side), by the Side - Angle - Side (SAS) congruence criterion, $\triangle ABD\cong\triangle ACD$.

Step2: Get $BD = CD$

Corresponding parts of congruent triangles are congruent. So, $\overline{BD}\cong\overline{CD}$ because they are corresponding sides of $\triangle ABD$ and $\triangle ACD$.

Step3: Prove $\angle DBC\cong\angle DCB$

In $\triangle BDC$, since $\overline{BD}\cong\overline{CD}$, by the Isosceles - Triangle Theorem (if two sides of a triangle are congruent, then the angles opposite those sides are congruent), $\angle DBC\cong\angle DCB$.

Answer:

The proof is completed as above to show that $\angle DBC\cong\angle DCB$.