Question 69

Sides AB and DC of cyclic quadrilateral ABCD are produced to meet at E, and sides AD and BCare produced to meet at F. If $$\angle BAD = 102^\circ$$ and $$\angle BEC = 38^\circ$$ then the difference between $$\angle ADC$$ and $$\angle AFB$$ is:

Solution

In ΔADE,
∠ADE=180 − (∠AED + ∠EAD)
= 180 − (38 + 102)
= 40$$\degree$$
⇒∠ADC = 40$$\degree$$
square ABCD is a cyclic quadrilateral.
∴∠DCB + ∠DAB=180
⇒∠DCB = 180 − ∠DAB
∠DCB = 180 − 102
∠DCB = 78$$\degree$$
In ΔDFC,
∠DFC=180 - (∠FDC+∠FCD)
∠DFC = 180 − (40 + 78)
∠DFC = 180 − 118
∠DFC = 62$$\degree$$
∠AFB = ∠DFC = 62$$\degree$$.