在等腰三角形ABC中,CH是底边上的高线,点P是线段CH上不与端点重合的的任意一点在等腰△ABC中,CH是底边上的高线,点P是线段CH上不与端点重合的任意一点 连接AP交BC于点E,连接BP交AC于点F.（1）证

在等腰三角形ABC中,CH是底边上的高线,点P是线段CH上不与端点重合的的任意一点在等腰△ABC中,CH是底边上的高线,点P是线段CH上不与端点重合的任意一点 连接AP交BC于点E,连接BP交AC于点F.（1）证
在等腰三角形ABC中,CH是底边上的高线,点P是线段CH上不与端点重合的的任意一点
在等腰△ABC中,CH是底边上的高线,点P是线段CH上不与端点重合的任意一点 连接AP交BC于点E,连接BP交AC于点F.（1）证明：∠CAE=∠CBF（2）证明：AE=BF （3）以线段AE,BF和AB为边构成一个新的△ABG,记△ABC和△ABG的面积分别为S△ABC和S△ABG,当∠ACB=100°和∠ACB=80°时,分别讨论是否存在点P,能使得S△ABC=S△ABG成立.

在等腰三角形ABC中,CH是底边上的高线,点P是线段CH上不与端点重合的的任意一点在等腰△ABC中,CH是底边上的高线,点P是线段CH上不与端点重合的任意一点 连接AP交BC于点E,连接BP交AC于点F.（1）证
（1）因为在等腰△ABC中AC=BC,CH是底边上的高线,所以∠ACH=∠BCH,即∠ACP=∠BCP.
因为∠CAE=∠CBF,即∠CAP=∠CBP,又因为AC=BC,CP=CP,所以△ACP和△BCP全等,所以∠CAP=∠CBP.
因为AC=BC,∠CAP=∠CBP,∠ACE=∠ACB=∠BCF,所以△ACE和△BCF全等,所以∠CAE=∠CBF.
（2）因为△ACE和△BCF全等,所以AE=BF.
（3）由于△ABC和△ABG同底,要使S△ABC=S△ABG,必须要使两者的高相等.假设△ABG在AB边上的高是GQ,则CH=GQ.
因为AE=BF,所以在等边△ABG中,AQ=BQ=1/2AB.
同样在等腰△ABC中,AH=BH=1/2AB.
因此AH=BH=AQ=BQ.
又因为∠AHC=∠BHC=∠AQC=∠BQC=90°,CH=GQ,所以△AHC,△BHC,△AQG和△BQG全等,所以AC=BC=AG=BG,即AC=BC=AE=BF.
所以∠ACE=∠ACB=∠AEC,∠BFC=∠ACB=∠BCF.
当∠ACB=100°,不存在一点P满足∠ACB=∠AEC=100°.
当∠ACB=80°,存在一点P满足∠ACB=∠AEC=80°.
因此,当∠ACB=100°时,不存在点P,能使得S△ABC=S△ABG成立；,当∠ACB=80°时,存在点P,能使得S△ABC=S△ABG成立.

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（1）因为在等腰△ABC中AC=BC,CH是底边上的高线,所以∠ACH=∠BCH, 即∠ACP=∠BCP。
因为∠CAE=∠CBF，即∠CAP=∠CBP，又因为AC=BC，CP=CP,所以△ACP和△BCP全等，所以∠CAP=∠CBP。
因为AC=BC,∠CAP=∠CBP,∠ACE=∠ACB=∠BCF,所以△ACE和△BCF全等，所以∠CAE=∠CBF。
（...

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（1）因为在等腰△ABC中AC=BC,CH是底边上的高线,所以∠ACH=∠BCH, 即∠ACP=∠BCP。
因为∠CAE=∠CBF，即∠CAP=∠CBP，又因为AC=BC，CP=CP,所以△ACP和△BCP全等，所以∠CAP=∠CBP。
因为AC=BC,∠CAP=∠CBP,∠ACE=∠ACB=∠BCF,所以△ACE和△BCF全等，所以∠CAE=∠CBF。
（2）因为△ACE和△BCF全等，所以AE=BF。
（3）由于△ABC和△ABG同底，要使S△ABC=S△ABG，必须要使两者的高相等。假设△ABG在AB边上的高是GQ，则CH=GQ。
因为AE=BF，所以在等边△ABG中，AQ=BQ=1/2AB。
同样在等腰△ABC中，AH=BH=1/2AB。
因此AH=BH=AQ=BQ。
又因为∠AHC=∠BHC=∠AQC=∠BQC=90°,CH=GQ,所以△AHC，△BHC,△AQG和△BQG全等，所以AC=BC=AG=BG，即AC=BC=AE=BF。
所以∠ACE=∠ACB=∠AEC,∠BFC=∠ACB=∠BCF。
当∠ACB=100°,不存在一点P满足∠ACB=∠AEC=100°。
当∠ACB=80°,存在一点P满足∠ACB=∠AEC=80°。
因此,当∠ACB=100°时,不存在点P,能使得S△ABC=S△ABG成立；,当∠ACB=80°时,存在点P,能使得S△ABC=S△ABG成立.

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