题目内容
(1)写出图中两对相似三角形(不得添加字母和线);
(2)请分别说明两对三角形相似的理由.
分析:(1)△ABC∽△ADE,△ABD∽△ACE;
(2)∠BAD=∠CAE,在此等式两边各加∠DAC,可证∠BAC=∠DAE,再结合已知中的∠ABC=∠ADE,可证△ABC∽△ADE;利用△ABC∽△ADE,可得AB:AD=AC:AE,再结合∠BAD=∠CAE,也可证△BAD∽△CAE.
(2)∠BAD=∠CAE,在此等式两边各加∠DAC,可证∠BAC=∠DAE,再结合已知中的∠ABC=∠ADE,可证△ABC∽△ADE;利用△ABC∽△ADE,可得AB:AD=AC:AE,再结合∠BAD=∠CAE,也可证△BAD∽△CAE.
解答:解:(1)△ABC∽△ADE,△ABD∽△ACE(2分)
(2)①证△ABC∽△ADE,
∵∠BAD=∠CAE,
∠BAD+∠DAC=∠CAE+∠DAC,
即∠BAC=∠DAE.(4分)
又∵∠ABC=∠ADE,
∴△ABC∽△ADE.(5分)
②证△ABD∽△ACE,
∵△ABC∽△ADE,
∴
=
.(7分)
又∵∠BAD=∠CAE,
∴△ABD∽△ACE.(8分)
(2)①证△ABC∽△ADE,
∵∠BAD=∠CAE,
∠BAD+∠DAC=∠CAE+∠DAC,
即∠BAC=∠DAE.(4分)
又∵∠ABC=∠ADE,
∴△ABC∽△ADE.(5分)
②证△ABD∽△ACE,
∵△ABC∽△ADE,
∴
| AB |
| AD |
| AC |
| AE |
又∵∠BAD=∠CAE,
∴△ABD∽△ACE.(8分)
点评:本题利用了等量加等量和相等、相似三角形的判定和性质.
练习册系列答案
相关题目