如图,已知AP平分∠CAD,BP平分∠CBD,两角平分线交于点P,且∠C=32°∠D=28°,求∠P的度数,

式子一：∠P+1/2∠B=1/2∠A+28°
式子二：∠P+1/2∠A=1/2∠B+32°
两式子相加得到：2∠P+1/2∠A+1/2∠B=1/2∠A+1/2∠B+28°+32°
2∠P=60°
∠P=30°
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