Question

如图,圆O的直径AB长为6,弦AC长为2,角ACB的平分线交圆O于点D，求弦CD的长

Answer 1

解：过点A作AE⊥CD
∵AB为圆O直径
∴∠ACB＝90, ∠ADB＝90
∵CD平分∠ACB
∴∠ACD＝∠BCD＝45
∵∠BCD、∠BAD所对应圆弧都为劣弧BD
∴∠BAD＝∠BCD=45
∵AB＝6
∴AD＝6×cos∠BAD=6×(√2/2)=3√2
∵AC=2
∴BC=√(AB²-AC²)=√(36-4)=4√2
∴cos∠ABC=BC/AB=4√2/6=2√2/3
sin∠ABC=AC/AB=2/6=1/3
∵∠ABC、∠ADC所对应圆弧都为劣弧AC
∴∠ADC＝∠ABC
∵AE⊥CD
∴DE＝AD×cos∠ADC＝AD×cos∠ABC＝3√2×2√2/3＝4
AE=AD×sin∠ADC=AD×sin∠ABC=3√2×1/3＝√2
∵AE⊥CD, ∠ACD＝45
∴CE=AE=√2
∴CD=CE+DE=√2+4