第1个回答 2010-08-27
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第2个回答 2012-04-12
（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.
第3个回答 推荐于2016-12-01
（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.本回答被提问者采纳