Question

如图,在梯形ABCD中,AD//BC,∠BAC=90°,AB=AC,BD=BC,AC,BD相交于O点,求证:OC=OD

Answer 1

求证:OC=CD

过A、D分别作AE⊥ BC、DF⊥ BC,垂足分别为E、F,则DF=AE=1/2BC=1/2BD

DF=1/2BD,DF⊥ BC=>∠DBC=30°,∵BD=BC=>∠BDC=∠BCD=1/2(180°-30°)=75°---(1)

∠OCD=∠BCD-∠BCA=75°-45°=30°=>∠DOC=180°-∠OCD-∠BDC=180°-30°-75°=75°----(2)

(1)(2)=>∠BDC=∠DOC=>OC=CD