In isosceles triangle ABC, AC ≅ BC, and CD is the altitude from C to base AB, with m∠ACD = 55°. Using triangle congruence and CPCTC, what is m∠BCD?
A55°, because ∠ACD and ∠BCD are vertical angles formed where CD crosses AB at D
B55°, because △ACD ≅ △BCD by HL (AC≅BC, CD≅CD, right angles at D), so ∠ACD ≅ ∠BCD by CPCTC
C110°, because ∠BCD is twice ∠ACD by CPCTC
D35°, because ∠ACD and ∠BCD must be complementary angles that together make up a right angle at the apex
Explanation
CD ⊥ AB creates two right angles at D. AC ≅ BC (given) and CD ≅ CD (Reflexive), so by HL, △ACD ≅ △BCD. By CPCTC, ∠ACD ≅ ∠BCD, so m∠BCD = m∠ACD = 55°.