1. Solution: Because BD and CD are bisectors of the two outer corners of △ABC.

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