In triangle $ABC$, $ \angle ABC = 2 \alpha $ and $ \angle ACB = 140 ^\circ - \alpha $. $D$ is a point on line segment $AB$ such that $ CB = BD $. What is the measure (in degrees) of $ \angle DCA $?
+26367 In triangle \(ABC\), \(\angle ABC = 2 \alpha\) and \(\angle ACB = 140 ^\circ - \alpha\).
\(D\) is a point on line segment \(AB\) such that \(CB = BD\).
What is the measure (in degrees) of \(\angle DCA\)?
\(\text{Let $\angle DCA=x$} \\ \text{Let $\angle BDC=\angle BCD= \beta$} \)
\(\begin{array}{|rcll|} \hline \mathbf{2\alpha + 2\beta} &=& \mathbf{180^\circ} \\ \alpha + \beta &=& 90^\circ \\ \mathbf{\beta} &=& \mathbf{90^\circ -\alpha} \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline \mathbf{140^\circ-\alpha} &=& \mathbf{x + \beta} \quad | \quad \mathbf{\beta=90^\circ -\alpha} \\ 140^\circ-\alpha &=& x + 90^\circ -\alpha \\ 140^\circ&=& x + 90^\circ \\ x &=& 140^\circ-90^\circ \\ \mathbf{x} &=& \mathbf{50^\circ} \\ \hline \end{array}\)
