Constructions Around a Circle

I approach problems like this by identifying special cases that make it easier to work w/. Here, the special case I use is when EF⊥CD:

1). Tagent-chord theorem => ∠AEC = ∠ADC = 25°.

2). ∆ACE ≌ ∆ACE => ∠BEA = ∠CEA = 25°; ∆ACE ≌ ∆ACE => AG ⊥ BC. Thus ∠CBE = 65°.

3). ∠CBE = ∠DBE & EF ⊥ CD => CF = DF, & hence ∠DEF = 50°.

Let $CD$ intersect $\Gamma$ again at $G$ . Let $EG$ extended intersect $AD$ at $H$ . By alternate segment theorem, $\angle AED = \angle ACD = \angle ACG = \angle GEC$ , so $\angle DEG = \angle CEA = \angle CDA = \angle 25^\circ$ .

Observe that $HDG$ and $HED$ are similar triangles, so $HD^2 = HG \times HE$ . By power of a point, the power of $H$ with respect to $\Gamma$ gives $HB^2 = HG \times HE$ . Hence $HD = HB$ , so $H$ is the midpoint of the hypotenuse, and $HD = HB = HF$ .

Hence, $\angle HFD = \angle HDF = 25^\circ = \angle HED$ , so points $D, E, F, H$ are concyclic. This shows that $\angle HEF = \angle HDF = 25^\circ$ . Thus, $\angle DEF = \angle DEH + \angle HEF = 50 ^\circ$ .

Let M be midpoint of BD, DC intersect circle with N, then BM=MD=MF=sqrt(power of M), MN must pass E by congruence