Question 2.6.13: In a convex pentagon ABCDE, AB = BC, ∠ABE + ∠DBC = ∠EBD, and......

By the assumption ∠AEB + ∠BDC = π, there exists a point F on AC such that ∠AFB = ∠AEB and ∠BFC = ∠BDC; this means F is the second intersection of the circumcircles of BCD and ABE. The triangle ABC is isosceles, so ∠FCB = (π − ∠ABC)/2. Hence ∠FDB = ∠FCB = π/2 − ∠DBE by the assumption ∠ABE + ∠DBC = ∠EBD, and so DF ⊥ BE. Similarly EF ⊥ BD, and so F is the orthocenter of BDE.