∠∠ACD =∠ECB = 90 (known)

∠∠ACD =∠ACE+∠ECD = 90。

∠ECB=∠ECD+∠BCD=90

∴∠ACE=∠BCD (equivalent substitution)

(2) solutions; ∠∠ACD =∠ACE+∠ECD = 90 (as can be seen from the above questions).

∫∠DCE = 30 (known)

∴∠ACE=∠ACD-∠ECD=90 -30 =60

∠∠ACB =∠ACE+∠ECB = 60+90 = 150

(3) solutions; ∠ACB and ∠DCE are complementary.

∠∠ACB = 150, ∠ DCE = 30 (known)

∴∠acb+∠dce= 150+30 = 180

∴∠ACB and ∠DCE are complementary.

(4) Established