如图，在三角形ABC中，角ABC和角ACB的外角平分线相交于点P。（1）若角ABC=30度，角ACB=70度，求角BPC的度

\u95ee: \u5982\u56fe,\u5728\u4e09\u89d2\u5f62ABC\u4e2d,\u89d2ABC\u548c\u89d2ACB\u7684\u5916\u89d2\u5e73\u5206\u7ebf\u76f8\u4ea4\u4e8e\u70b9P.\u82e5\u89d2ABC=\u03b1,\u89d2BPC=

\u8fd9\u9053\u9898\u7528\u7684\u77e5\u8bc6\u662f1.\u4e09\u89d2\u5f62\u5185\u89d2\u548c\u4e3a180\u5ea6\uff0c2.\u8fb9\u7ec4\u6210\u4e00\u76f4\u7ebf\u7684\u591a\u4e2a\u89d2\u7684\u548c\u4e3a180\u5ea6\uff08\u76f4\u7ebf\u662f180\u5ea6\u89d2\uff09\u3002
\u9898\u76ee\u6d89\u53ca\u5230\u4e24\u4e2a\u4e09\u89d2\u5f62\uff0c\u4f46\u7531\u4e8e\u4e09\u89d2\u5f62ABC\u53ea\u77e5\u9053\u4e00\u4e2a\u89d2\uff0c<ACB\u662f\u8981\u6c42\u51fa\u7684\uff0c<BAC\u4e0d\u53ef\u80fd\u901a\u8fc7\u4efb\u4f55\u5173\u7cfb\u77e5\u9053\u3002
\u6211\u4eec\u53ea\u80fd\u901a\u8fc7\u4e09\u89d2\u5f62BPC\u6c42\u5f97\uff1a
<CBP=(180-\u03b1)/2
<BCP=180-<CBP-<BPC=180-(180-\u03b1)/2-\u03b2
<ACB=180-2\u00d7<BCP=180-\uff08360-180+\u03b1-2\u03b2\uff09=2\u03b2-\u03b1

\u5443\u00b7~ \u5e94\u8be5\u662f\u8fd9\u6837\uff0c\u81f3\u5c11\u65b9\u6cd5\u662f\u5bf9\u7684

\uff081\uff09\u2220BPC=180\u00b0-\uff0812\u2220EBC+12\u2220BCF\uff09=180\u00b0-12\uff08\u2220EBC+\u2220BCF\uff09=180\u00b0-12\uff08180\u00b0-\u2220ABC+180\u00b0-\u2220ACB\uff09=180\u00b0-12\uff08180\u00b0-30\u00b0+180\u00b0-70\u00b0\uff09=50\u00b0\uff1b\uff082\uff09\u2220BPC=180\u00b0-12\uff08180\u00b0-\u2220ABC+180\u00b0-\u2220ACB\uff09=12\uff08\u2220ABC+\u2220ACB\uff09\uff0c\u2235\u2220BPC=\u03b2\uff0c\u2220ABC=\u03b1\uff0c\u2234\u03b2=12\uff08\u03b1+\u2220ACB\uff09\uff0e\u6545\u2220ACB=2\u03b2-\u03b1\uff0e

解：
(1)∵∠B和∠C的外角平分线相交于点P，
∴∠PBC=1/2∠EBC, ∠PCB=1/2∠FCB.
∵∠A=80°，∴∠ABC+∠ACB= 100°.
又∵∠ABC +∠ACB+∠EBC+∠FCB=360°，
∴∠EBC+∠FCB= 360°．
又∵∠PBC+∠PCB+∠BPC= 180°．
∴∠BPC = 180° - (∠PBC +∠PCB ) = 180°-1/2 (∠EBC+∠FCB)=50°

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