91Ó°ÊÓ

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

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

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

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

Therefore, if$m∠C=22$ then $m∠ABD=22$.

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

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

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

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

Therefore,$m∠C=23$ if then $m∠ABD=23$.

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

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

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

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

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

Substitute$90−m∠C$ for$m∠CAB$ into the equation obtained in step 3.

$$\begin{gathered} 90−m∠C+90+m∠ABD=180 \\ m∠ABD−m∠C=180−180 \\ m∠ABD=m∠C \end{gathered}$$

In$\mathrm{Δ}ABC$ and $\mathrm{Δ}ABD$,$∠ABCâ‰\dots ∠ADB$ and$∠BACâ‰\dots ∠BAD$ then by corollary 1 $m∠ABD$is always equal to $m∠C$.