In a circle, chords AD and BC meet at a point E outside the circle. If ∠BAE = 76° and ∠ADC = 102°, then ∠AEC is equal to?

For secants intersecting outside a circle: ∠AEC = (∠ADC - ∠BAE)/2. Here, ∠AEC = (102° - 76°)/2 = 26°/2 = 13°. Wait, let me reconsider. Actually, for two chords AD and BC meeting at E outside the circle, the angle formed equals half the difference of the intercepted arcs. Or, using the property: ∠AEC = 180° - ∠BAE - ∠ABE. And ∠ABE = ∠ADC (angles subtended by same arc AC). So ∠AEC = 180° - 76° - 102° = 2°, which doesn't match. Let me use the secant theorem: For point E outside, ∠AEC = (∠BEC - ∠BAE)... Actually the correct formula is: ∠AEC = 1/2(|arc(AC) - arc(BD)|). Given ∠BAE = 76° and ∠ADC = 102°, we need to find which arcs these intercept. Using the exterior angle theorem for cyclic quadrilaterals and properties of secants, the answer is 26°.