Quadrilateral \(ACDB\) has \(AC=BC=DC\), \(\angle ADB=30\degree\), and \(\angle CAD=10\degree\).
Determine the measure of \(\angle ACB\).
Since \(AC=DC\), \(\triangle ACD\) is isosceles. Therefore, \(\angle CDA = \angle CAD = 10\degree\).
Thus, \(\angle CDB = \angle CDA + \angle ADB = 10\degree + 30\degree = 40\degree\).
Since \(BC = DC\), \(\triangle BCD\) is isosceles. Therefore, \(\angle CBD = \angle CDB = 40\degree\).
Since the angles in a triangle sum to \(180\degree\), in \(\triangle ACD\) we have \[\begin{aligned} \angle CAD + \angle CDA + \angle ACD &= 180\degree\\ 10\degree + 10\degree + \angle ACD &= 180\degree\\ 20\degree + \angle ACD &= 180\degree\\ \angle ACD &= 160\degree \end{aligned}\]
Since the angles in a triangle sum to \(180\degree\), in \(\triangle BCD\) we have \[\begin{aligned} \angle CDB + \angle CBD + \angle BCD &= 180\degree\\ 40\degree + 40\degree + \angle BCD &= 180\degree\\ 80\degree + \angle BCD &= 180\degree\\ \angle BCD &= 100\degree \end{aligned}\]
Since \(\angle ACD = \angle ACB + \angle BCD\), we have \(160\degree = \angle ACB + 100\degree\). Therefore, \(\angle ACB = 160\degree - 100\degree = 60\degree\).
Extension:
Suppose \(\angle ADB=30\degree\) and \(\angle CAD=x\degree\). Show that \(\angle ACB = 60\degree\).