Sides AB and DC of a cyclic quadrilateral ABCD are produced to meet at E and sides AD and BC are produced to meet at F. If ∠ADC = 78° and ∠BEC = 52°, then the measure of ∠AFB is:
Answer & explanation
Correct answer: option 1
\(\angle\)ADC = \({78}^\circ\) and \(\angle\)BEC = \({52}^\circ\)
As we know, in a cyclic quadrilateral, the sum of opposite angles are \({180}^\circ\).
\(\angle\)ADC + \(\angle\)ABC = 180
= \(\angle\)ABC = 180 - 78 = 102
= \(\angle\)ABC + \(\angle\)CBE = 180
= \(\angle\)CBE = 180 - 102 = 78
In \(\Delta \)BEC
= \(\angle\)CBE + \(\angle\)BEC + \(\angle\)ECB = 180
= \(\angle\)ECB = 180 - 78 - 52 = 50
= \(\angle\)ECB + \(\angle\)BCD = 180
= \(\angle\)BCD = 180 - 50 = 130
= \(\angle\)BAD + \(\angle\)BCD = 180
= \(\angle\)BAD = \(\angle\)BAF = 180 - 130 = 50
In \(\Delta \)AFB
= \(\angle\)BAF + \(\angle\)ABF + \(\angle\)AFB = 180
= \(\angle\)AFB = 180 - 50 - 102 = 28
Therefore, \(\angle\)AFB is \({28}^\circ\).