Similar Right Triangles

$\newenvironment {prompt}{}{} \newcommand {\ungraded }[0]{} \newcommand {\framed }[0]{\MakeFramed {\advance \hsize -\width \FrameRestore }} \newcommand {\shaded }[0]{\def \FrameCommand {\fboxsep =\FrameSep \colorbox {shadecolor}}\MakeFramed {\FrameRestore }} \newcommand {\shaded* }[0]{\def \FrameCommand {\fboxsep =\FrameSep \colorbox {shadecolor}}\MakeFramed {\advance \hsize -\width \FrameRestore }} \newcommand {\leftbar }[0]{\def \FrameCommand {\vrule width 3pt \hspace {10pt}}\MakeFramed {\advance \hsize -\width \FrameRestore }} \newcommand {\titled-frame }[1]{\def \FrameCommand {\fboxsep 8pt\fboxrule 2pt \TitleBarFrame {\textbf {#1}}}\def \FirstFrameCommand {\fboxsep 8pt\fboxrule 2pt \TitleBarFrame [$\blacktriangleright $]{\textbf {#1}}}\def \MidFrameCommand {\fboxsep 8pt\fboxrule 2pt \TitleBarFrame [$\blacktriangleright $]{\textbf {#1\ (cont)}}}\def \LastFrameCommand {\fboxsep 8pt\fboxrule 2pt \TitleBarFrame {\textbf {#1\ (cont)}}}\MakeFramed {\advance \hsize -20pt \FrameRestore }} \newcommand {\FrameCommand }[0]{\setlength \fboxrule {\FrameRule }\setlength \fboxsep {\FrameSep }\fbox } \newcommand {\FirstFrameCommand }[0]{\FrameCommand } \newcommand {\MidFrameCommand }[0]{\FrameCommand } \newcommand {\LastFrameCommand }[0]{\FrameCommand } \newcommand {\FrameHeightAdjust }[0]{6pt} \newcommand {\sectionOutcomes }[0]{} \newcommand {\N }[0]{\mathbb N} \newcommand {\W }[0]{\mathbb W} \newcommand {\C }[0]{\mathbb C} \newcommand {\Z }[0]{\mathbb Z} \newcommand {\Q }[0]{\mathbb Q} \newcommand {\R }[0]{{\mathbb R}} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument }$

Two proofs.

Adapted from Ohio’s 2017 Geometry released item 17.

Complete the following proof that $\triangle DAC$ is similar to $\triangle DBA$ :

(a): $\triangle DBA\sim \triangle \answer [format=string]{ABC}$ by because they share $\angle B$ and they each have a right angle.
(b): $\triangle DAC\sim \triangle \answer [format=string]{ABC}$ for the same reason because they share $\angle A$ $\angle B$ $\angle C$ and they each have a right angle.
(c): $\triangle DAC\sim \triangle DBA$ because CPCTCright triangle similaritythey are both similar to $\triangle ABC$ .

A different proof, also adapted from Ohio’s 2017 Geometry released item 17.

Complete the following proof that $\triangle DAC$ is similar to $\triangle DBA$ :

(a): $\angle B$ and $\angle BAD$ are $\answer [format=string]{complementary}$ because they are acute angles in a right triangle.
(b): $\angle DAC$ and $\angle BAD$ are complementary because they are adjacent angles that form $\angle BAC$ , which is rightacuteobtuse .
(c): $\angle B \cong \angle DAC$ because they are both complementary to $\angle BAD$ .
(d): $\triangle DAC\sim \triangle DBA$ by because $\angle B \cong \angle DAC$ and they each have a right angle.