ABCD is a cyclic quadrilateral such that when sides AB and DC are produced, they meet at E, and sides AD and BC meet at F, when produced. If ∠ADE = 80° and ∠AED = 50°, then what is the measure of ∠AFB ?

30°

From the above diagram,

= \(\angle\)ADC + \(\angle\)CDF = 180     [Linear pair of angles]

= 80 + \(\angle\)CDF = 180

= \(\angle\)CDF = 100

Now, In \(\Delta \)AED

= \(\angle\)AED + \(\angle\)ADE + \(\angle\)EAD = 180

= 50 + 80 + \(\angle\)EAD = 180

= \(\angle\)EAD = 180 - 130 = 50

Now, \(\angle\)EAD = \(\angle\)DCF = 50

In \(\Delta \)DCF

= \(\angle\)CDF + \(\angle\)DFC + \(\angle\)DCF = 180

= 100 + \(\angle\)DFC + 50 = 180

= \(\angle\)DFC = 180 - 150 = 30

= \(\angle\)DFC = \(\angle\)AFB = \({30}^\circ\)

Therefore, \(\angle\)AFB is \({30}^\circ\).