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設∠B=x,∵∠DCB=2∠B,∴∠DCB=2x,∵∠C的平分線交AB於D,∴∠ACD=∠DCB=2x,∵∠ADC是△BCD的外角,∴∠ADC=∠B+∠DCB=3x,在△ACD中,∵∠A+∠ACD+∠ADC=180°,∴90°+2x+3x=180°,解得x=18°,∴∠ADC=3x=3×18°=54°.
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