Given:$\overline{\mathit{A}\mathit{B}}\perp \overline{\mathit{B}\mathit{C}};\overline{\mathit{B}\mathit{D}}\perp \overline{\mathit{A}\mathit{C}}$

1. If $m\angle C=22$, find $m\angle ABD$.
2. If $m\angle C=23$, find_{$m\angle ABD$}.
3. Explain why $m\angle ABD$ always equal$m\angle C$.
   
   

Consider a $\mathrm{\Delta }ABC$, such that,

$$\begin{gathered} m\angle CAB+m\angle CBA+m\angle BCA=180 \\ m\angle CAB+90+22=180 \\ m\angle CAB=180-112 \\ =68 \end{gathered}$$

Consider a $\mathrm{\Delta }ABD$, such that,

$$\begin{gathered} m\angle BAD+m\angle ADB+m\angle ABD=180 \\ 68+90+m\angle ABD=180 \\ m\angle ABD=180-158 \\ =22 \end{gathered}$$

Therefore, if$m\angle C=22$ then $m\angle ABD=22$.

Consider a $\mathrm{\Delta }ABC$, such that,

$$\begin{gathered} m\angle CAB+m\angle CBA+m\angle BCA=180 \\ m\angle CAB+90+23=180 \\ m\angle CAB=180-113 \\ =67 \end{gathered}$$

Consider a $\mathrm{\Delta }ABD$, such that,

$$\begin{gathered} m\angle BAD+m\angle ADB+m\angle ABD=180 \\ 67+90+m\angle ABD=180 \\ m\angle ABD=180-157 \\ =23 \end{gathered}$$

Therefore,$m\angle C=23$ if then $m\angle ABD=23$.

The sum ofmeasures ofall interior angles of a triangle is$180°$.

Consider a $\mathrm{\Delta }ABC$, such that,

$$\begin{gathered} m\angle CAB+m\angle CBA+m\angle BCA=180 \\ m\angle CAB+90+m\angle BCA=180 \\ m\angle CAB+m\angle BCA=180-90 \\ m\angle CAB+m\angle C+=90 \\ m\angle CAB=90-m\angle C \end{gathered}$$

Consider a $\mathrm{\Delta }ABD$, such that,

$\begin{array}{c}m\angle BAD+m\angle ADB+m\angle ABD=180\\ m\angle CAB+m\angle ADB+m\angle ABD=180\end{array}$

Substitute$90-m\angle C$ for$m\angle CAB$ into the equation obtained in step 3.

$$\begin{gathered} 90-m\angle C+90+m\angle ABD=180 \\ m\angle ABD-m\angle C=180-180 \\ m\angle ABD=m\angle C \end{gathered}$$

In$\mathrm{\Delta }ABC$ and $\mathrm{\Delta }ABD$,$\angle ABC\cong \angle ADB$ and$\angle BAC\cong \angle BAD$ then by corollary 1 $m\angle ABD$is always equal to $m\angle C$.