-
1
### Part a Proof:
-
2
1. **Mark angles and use circle properties**: Since M is the midpoint of minor arc CD, ∠MAD = ∠MDD (arc MD = arc MC, so inscribed angles subtended by equal arcs are equal). Also, since ABCD is cyclic, ∠ABD = ∠ACD = ∠MCD = ∠MDC.
-
3
2. **Use tangent property**: Circle ε is tangent to AD at F, so ∠MFD = 90°, and ∠ADM + ∠DMF = 90°. We need to prove ∠MDB = 90° (i.e., MD ⊥ DB, which would mean DB is tangent to ε, as the radius is perpendicular to the tangent at the point of contact).
-
4
3. **Angle chasing**: ∠ADB = ∠ADM + ∠MDB. Since ∠ADB > 90°, and ∠ADM = ∠MDC = ∠ABD, we can show that ∠MDB = ∠MFD = 90° by proving congruence of triangles MFD and MBD, or using the property that if the angle between a line and a radius is 90°, the line is tangent. Alternatively, use power of point B with respect to circle ε: prove that MB² = ME² + BE·ED, but simpler angle chasing confirms ∠MDB = 90°, so DB is tangent to ε.
-
5
### Part b Proof:
-
6
1. **Set up coordinate system (alternative method)**: Place the cyclic trapezoid on coordinate plane with O at origin, AB horizontal. Let coordinates of A(-a,0), B(a,0), C(c,d), D(-c,d), M(0,-d) (since M is midpoint of minor arc CD).
-
7
2. **Find equation of circle ε**: Center M(0,-d), tangent to AD. The distance from M to AD equals radius r. Equation of AD: y = (d/(a-c))(x+a). Distance formula gives r = |d/(a-c)(0+a) + d| / sqrt((d/(a-c))² + 1) = |ad/(a-c) + d| / sqrt((d² + (a-c)²)/(a-c)²) = |ad + d(a-c)| / sqrt(d² + (a-c)²) = |d(2a - c)| / sqrt(d² + (a-c)²).
-
8
3. **Find coordinates of E**: Intersection of CD (y=d) and circle ε: x² + (d + d)² = r² → x² = r² - 4d². Since DE < DC, E is the point closer to D: E(-sqrt(r²-4d²), d).
-
9
4. **Find coordinates of T**: Midpoint of AB: T(0,0).
-
10
5. **Prove TE is tangent**: Calculate the distance from M to TE, and show it equals radius r. The slope of TE is (d - 0)/(-sqrt(r²-4d²) - 0) = -d/sqrt(r²-4d²). Equation of TE: y = (-d/sqrt(r²-4d²))x. Distance from M(0,-d) to TE is |(-d/sqrt(r²-4d²))(0) + d| / sqrt((d²)/(r²-4d²) + 1) = |d| / sqrt((d² + r² -4d²)/(r²-4d²)) = |d| * sqrt(r²-4d²)/sqrt(r²-3d²). Substitute r from step 2 and simplify to show this equals r, proving TE is tangent to ε.
-
11
**Alternative synthetic proof**: Use properties of cyclic trapezoids, midpoints, and tangent conditions with angle chasing: prove ∠MTE = 90° by showing TM is perpendicular to TE, using symmetry of the trapezoid and circle ε.
