Question

如图在梯形ABCD中ad平行bc,∠abc等于90°,ab=ac,bd=bc,abcd相交于o点,求证oc=cd

Answer 1

作AE⊥BC,DF⊥DC,∵AB=AC,∠BAC=90°,∠ABC=∠ACB=45°,∴AE=DF=1/2BC,又∵BC=BD,∴DF=1/2BD,即∠DAF=30°,∴∠BCD=∠BDC=75°,∠OCD=∠BCD-∠ACD=30°,∴∠COD=75°,即OC=CD