⒈如图一,在锐角△ABC中,CD垂直于AB于点D,E是AB上的一点.找出图中所有的锐角三角形,并说明理由.图见:
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/0eb30f2442a7d9337aa0e640ad4bd11373f001fc?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
⒉如图二,△ABC中,∠B大与∠C,AD是∠BAC的平分线,说明∠ADB-∠ADC=∠C-∠B成立的理由.图见:
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/43a7d933c895d143197ee61073f082025aaf07fc?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
⒊如图三,已知BO平分∠CBA,CO平分∠ACB,MN‖BC,AB=12,AC=18,求△AMN的周长.图见:
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/d833c895d143ad4b642644f382025aafa40f06fc?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
⒋如图四,已知△ABC中,AD是BC边上的高线,AE是∠BAC的平分线,若设∠EAD=a,求∠C-∠B.(用a的代数式表示)图见:
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/4afbfbedab64034fc6c6c035afc379310a551dfd?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
⒌如图五,已知AB=AC,AD=AE,∠1=∠2,问CE=BD吗?说明理由.图见:
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/024f78f0f736afc3b8073d56b319ebc4b74512fd?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
⒍如图六,由正方形ABCD边BC、CD向外作等边三角形BCE和CDF,连结AE、AF、EF,求证:△AEF为等边三角形。图见:
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/aec379310a55b3192ff2804643a98226cffc17fd?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
第一题:图一中共有三角形6个,为△ABC,△AEC,△CED,△CBD,△ACD,△ECB其中△CED,△ACD,△CDB为Rt△△AEC为钝角△,因为∠AEC=∠ADC+∠ECD=90°+∠ECD>90°△ABC锐角△,已知条件。∠CEB = 180°-钝角=锐角∠B为锐角,∠ECB=∠ACB-∠ACE =锐角△ECB为锐角△共有两个锐角△,为△ECB和△ACB第二题:∵AD是∠BAC的平分线∴∠BAD=∠DAC∵三角形内角和为180°∴∠BAD+∠B+∠ADB=∠DAC+∠ADC+∠C∴∠B+∠ADB=∠ADC+∠C∴∠ADB-∠ADC=∠C-∠B第三题∵MN‖BC∴∠MOB=∠OBC∴∠NOC=∠OCB∵BO平分∠CBA∴∠MBO=∠OBC∵CO平分∠ACB∴∠NCO=∠OCB∴∠MOB=∠MBO∴∠NCO=∠OCB∵∠MOB=∠MBO∴BM=OM∵∠NCO=∠OCB∴ON=NC∴AM+MN+NA = (AM+BM)+(AN+CN)=AB+AC=12+18=30∵△AMN的周长 = 30第四题∠C=90°-∠DAC = 90°-[(1/2)∠BAC-a]∠B=∠AEC-∠BAE = 90°- a-∠BAE = 90°- a-(1/2)∠BAC∠C-∠B=90°-[(1/2)∠BAC-a]-{90°- a-(1/2)∠BAC}=2a第六题∵正方形ABCD∴AB=AD=BC=CD∵△CDF和△BCE为等边△∵FD=DC,∴BE=AB,∴FD=BE∵∠ADF=∠ADC+∠FDC=90+60=150∵∠ABE=∠ABC+∠CBE=90+60=150∴∠DFA=∠DAF=∠BAE=∠BEA=15∴∠ADF=∠ABE∴△ADF≌△ABE∴AF=AE∴△AFE为等腰三角形∵∠FAE = ∠DAB-∠DAF-∠EAB =90°-15°-15°=60°∴△AFE为等边三角形