Line segments \(AB\) and \(CD\) are parallel, with \(AB\) above \(CD\). Point \(J\) lies above \(AB\), and points \(E\) and \(F\) lie on \(CD\), with \(E\) to the left of \(F\), so that \(JE\) intersects \(AB\) at \(G\) and \(JF\) intersects \(AB\) at \(H\).
If \(\angle CEG = 110\degree\) and \(\angle GHF = 122\degree\), determine the measure of \(\angle GJH\).
Solution 1
Since \(JHF\) is a straight line, then \(\angle JHG=180\degree-\angle GHF = 180\degree - 122\degree = 58\degree.\) Since \(AB\) and \(CD\) are parallel, \(\angle AGJ = \angle CEG = 110\degree.\) Since \(AGH\) is a straight line, \(\angle JGH = 180\degree - \angle AGJ = 180\degree -110\degree = 70\degree.\) Since the three angles in a triangle add to \(180\degree\), then \[\angle GJH = 180\degree - \angle JGH - \angle JHG = 180\degree - 70\degree - 58\degree = 52\degree.\]
Solution 2
Since \(JHF\) is a straight line, then \(\angle JHG=180\degree-\angle GHF = 180\degree - 122\degree = 58\degree.\) Since \(AB\) and \(CD\) are parallel, \(\angle JFE = \angle JHG = 58\degree.\) Since \(CEF\) is a straight line, \(\angle JEF = 180\degree - \angle CEG = 180\degree - 110\degree = 70\degree.\) Since the three angles in a triangle add to \(180\degree\), then \[\angle GJH = \angle EJF = 180\degree - \angle JEF - \angle JFE = 180\degree - 70\degree - 58\degree = 52\degree.\]
Note: Since \(\triangle GJH\) and \(\triangle EJF\) have all three angles in common, we can say that they are similar triangles. Similar triangles have properties that make them very useful in geometry problems.