In △ ABC, Ba bisects ∠ DBC, ∠ BAC = 118 ° BD ⊥ AC in D, and calculates the degree of ∠ C

In △ ABC, Ba bisects ∠ DBC, ∠ BAC = 118 ° BD ⊥ AC in D, and calculates the degree of ∠ C

∵∠BAC=∠D+∠DBA,∠BAC=118°,∠D=90°∴∠DBA=28°
∵ Ba bisection ∠ DBC, ∵ DBC = 2 ∠ DBA = 2 × 28 = 56 °

In △ ABC, Ba bisects ∠ DBC, ∠ BAC = 144 ° BD ⊥ AC in D, and calculates the degree of ∠ C
I always feel that this problem is wrong, and the degree of ∠ C is negative

Again, the topic should be wrong

Triangle ABC and triangle CDE are equilateral triangles, a, e and D are on the same line, and

Triangle AEC is equal to triangle BDC (AC = BC, CD = CE ∠ ACB = ∠ ECD ∠ ace = ∠ DCB)
Therefore, CAE = CBD
∠AEB=180-∠EAB-∠EBA
=180-(60-∠EAC)-(60-∠EBC)
=180-120+∠EBC+∠CBD
=60+62
=122

As shown in figure (1), ABC and CDE are equilateral triangles
1) Try to find the relationship between AE and BD
(2) If △ CDE is turned clockwise around point C to figure (2), is the above conclusion still true? Please explain the reason

No picture, only the first question
Because △ ABC △ CDE is equilateral
So in △ BCD and △ ACB
AC=BC,DC=EC
In addition, ACB = ACD = DCE = 60
Therefore, BCD = ace = 120
So △ BCD ≌ ACB
AE=BD