Join CA and BD, we know that angle subtended by arc at center of the circle is double the angle at remaining part by same arc
Thus ∠AOC = 2∠ADC
= 2 × 25°
∠AOC = 50°
Now AB || CD and AO is transversal. We know that line AO, intersects CD and AB at O and A respectively such that ∠OAB = ∠AOC and ∠OAB = 50°
In ∆OAB
OA = OB (radius of circle)
⇒ ∠OAB = ∠OBA(In a triangle angles opposite to equal sides arc equal)
⇒ ∠OBA = ∠OAB
⇒ ∠OBA = 50°
in ∆OAB
∠OAB + ∠OBA + ∠AOB = 180° [∵ Sum of triangle angles of a triangle ∆ is 180°]
50° + 50° + ∠AOB = 180°
100° + ∠AOB = 180°
∠AOB = 180°- 100°
∠AOB = 80°
Again, we know that angle subtended by an arc at center is double the angle at remaining part of circle by same arc.
∴ ∠AOB = 2∠AEB
⇒ 80° = 2∠AEB
⇒ ∠AEB = \(\frac { { 80 }^{ \circ } }{ 2 } \) = 40°
