question_answer

CD intersects the line AB at F, \[\angle \]CFB = \[50{}^\circ \] and \[\angle \]EFA = \[\angle \]AFD. Find the measure of \[\angle \]EFC.

Answer:

Let    \[\angle \]EFA = x Then  \[\angle \]AFD = x It is given that CD interests line AB at F. Therefore, \[\angle \]CFB - \[\angle \]AFD (vertically opposite angles) So,       x = 50° But \[\angle \]EFA = \[\angle \]AFD, which gives \[\angle \]EE4 ? 50 1 Now \[\angle \]CFB + \[\angle \]EFA + \[\angle \]EFC - 180° [as AB is a straight line] or, \[50{}^\circ +50{}^\circ \text{ }4-\]\[\angle \]EFC \[-\text{ }180{}^\circ \] or,  \[\angle \]EFC \[=-\text{ }180{}^\circ \text{ }-\text{ }100{}^\circ \] Thus, \[\angle \]EFC \[=80{}^\circ \]