So ∠CBD=∠DBE ∠BCD=∠DCF.
Let BCD = ∠ DCF = ∠ 1 ∠ BCD = ∠ DCF = ∠ 2.
The sum of the internal angles of a triangle is 180.
The inner and outer angles of a triangle are complementary.
free
∠ABC+∠ACB+∠A= 180
∠ABC+2∠ 1= 180
arrange
∠A+∠ACB=2∠ 1
In the same way.
∠A+∠ABC=2∠2
∠ 1+∠ 2 = ( 180-∠ A)/2
∠ BDC+∠ 1+∠ 2 = 180。
So ∠ BDC = 90+∠ A/2.
2 solution:
Ibid.: ∠ Abd = ∠ DBC = ∠1∠ ACD = ∠ DCE = ∠ 2.
∠A+2∠ 1=2∠2
∠ 1+∠BDC=∠2
Available < BDC = < A/2