Question

如图,△ABC中,D是BC延长线上一点,∠ABC和∠ACD的平分线相交于点P,∠A=70°,求∠P的度数.

Answer 1

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∠DCA = ∠A + ∠ABC (三角形外角等于不相邻内角和)

∠DCA

= 2(∠DCP) (角平分线)

= 2(∠P + ∠PBC) (三角形外角等于不相邻内角和)

= 2∠P + 2∠PBC

= 2∠P + ∠ABC (角平分线)

∠P = 1/2∠A = 35 (对比上下两条等式可得)