Question

it is given that $\triangle abc$ and $\triangle def$ are both right triangles and have exactly one right angle. all right angles are congruent, so
it is given that $\angle b\cong\angle e$ and that $\overline{ab}\cong d$
$\overline{ab}$ is
side between $\angle b$ ar
$\overline{de}$ is
side between $\angle e$ a
through the congruencies of
$,\angle b\cong\angle e$, and $\overline{ab}\cong\overline{de}$, it
follows that $\triangle abc\cong\triangle def$ by the
triangle congruency
theorem.

Explanation:

Step1: Analyze the right - angle congruence

Since \(\triangle ABC\) and \(\triangle DEF\) are right - triangles, assume \(\angle C = 90^{\circ}\) and \(\angle F=90^{\circ}\). So \(\angle C\cong\angle F\) (all right - angles are congruent).

Step2: Recall the ASA (Angle - Side - Angle) congruence criterion

For \(\triangle ABC\) and \(\triangle DEF\), if we have two angles and the included side congruent. In \(\triangle ABC\), \(\overline{AB}\) is the included side between \(\angle B\) and \(\angle C\). In \(\triangle DEF\), \(\overline{DE}\) is the included side between \(\angle E\) and \(\angle F\). Given \(\angle B\cong\angle E\), \(\overline{AB}\cong\overline{DE}\), and \(\angle C\cong\angle F\) (from Step 1), by the ASA (Angle - Side - Angle) Triangle Congruency Theorem, \(\triangle ABC\cong\triangle DEF\)

Answer:

The first drop - down (the congruent angles): \(\angle C\cong\angle F\)
The second drop - down (the type of side for \(\overline{AB}\)): included
The third drop - down (the type of side for \(\overline{DE}\)): included
The fourth drop - down (the congruent angles for the ASA criterion): \(\angle C\cong\angle F\)
The fifth drop - down (the congruency theorem): ASA