Question

Given:$\Delta \mathbf{ABC}$ in plane M; D not in plane M;

$\angle \mathbf{ACB}\cong \angle \mathbf{ABC};\angle \mathbf{DCB}\cong \angle \mathbf{DBC}$

Name a pair of congruent triangles. Prove that your answer is correct.

Short answer

A pair of congruent triangles is$\mathit{\Delta }\mathbf{ACD}$and $\Delta \mathbf{ABD}$.

Step-by-step solution

– Description of step.

If two angles of triangle are congruent then the sides opposite to those angles are congruent.

It is given that,$\angle ACB\cong \angle ABC$. Therefore, by converse isosceles triangle theorem,$\overline{AB}\cong \overline{AC}$.

– Description of step.

If two angles of triangle are congruent then the sides opposite to those angles are congruent.

It is given that,$\angle DCB\cong \angle DBC$. Therefore, by converse isosceles triangle theorem,$\overline{DB}\cong \overline{DC}$.

– Reflexive property.

Reflexive property states that, $\overline{PQ}\cong \overline{PQ}$. Therefore, in this case by reflexive property, $\overline{AD}\cong \overline{AD}$.

– SSS postulate.

If three sides of one triangle are congruent to three sides of other triangle then the triangles are congruent.

Here, $\overline{AB}\cong \overline{AC}$, $\overline{DB}\cong \overline{DC}$and $\overline{AD}\cong \overline{AD}$then by $SSS$postulate $\Delta ACD\cong \Delta ABD$.

Therefore, a pair of congruent triangles is$\Delta ACD$and $\Delta ABD$.