内容正文:
第四章 图形的相似
4.7 相似三角形的性质
第 1 课时
学习目标
1.经历探索相似三角形性质的过程,进一步体验由特殊到一般的归纳思想和方法,感悟转化的思想,积累数学活动经验.
2.了解相似三角形的性质定理:相似三角形对应高的比、对应角平分线的比、对应中线的比都等于相似比.
【教学目标... text has been truncated due to evaluation version limitation.
如果△ABC∽△A'B'C',由相似的定义,我们可以得到它们的边、角之间存在什么样的关系?
如果△ABC∽△A'B'C',相似比为k,
那么
∠A=∠A′,∠B=∠B′,∠C=∠C′.
复习引入
相似三角形的高、中线、角平分线之间关系?
【复习引入... text has been truncated due to evaluation version limitation.
3
如图,小王依据图纸上的△ABC,以1∶2的比例建造了模型房的房梁△A'B'C',CD和C'D'分别是它们的立柱.
(1)△ACD与△A'C'D'相似吗?为什么?
如果相似,指出它们的相似比.
(2)如果CD=1.5 cm,那么模型房的房梁立柱有多高?
D′
C′
B′
A′
探究新知
【探究新知... text has been truncated due to evaluation version limitation.
4
理由:∵ ,
∴△ABC∽△A'B'C'.
∴∠A=∠C'A'D'.
又∵CD⊥AB,
∴C'D'⊥A'B'.
∴∠ADC=∠A'D'C'=90°.
∴△ACD∽△A'C'D',且相似比为1∶2.
解:(1)△ACD与△A'C'D'相似;
探究新知
D′
C′
B′
A′
学生完成解... text has been truncated due to evaluation version limitation.
5
(2)∵△ACD∽△ A'C'D' ,
答:模型房的房梁立柱高3 cm.
∴C'D'=3.
∴ ,即
D′
C′
B′
A′
探究新知
设计意图:... text has been truncated due to evaluation version limitation.
6
想一想 已知△ABC∽△A'B'C',△ABC与△A'B'C'的相似比为k,它们对应高的比是多少?对应角平分线的比是多少?对应中线的比呢?请证明你的结论.
解:(1)由上面的“练一练”可得它们对应高的比是k.猜想:相似三角形对应高的比等于相似比.
探究新知
师生活动:... text has been truncated due to evaluation version limitation.
7
∵△ABC∽△A'B'C',
∴∠B=∠B'.
又∵△ABD和△A'B'D'都是直角三角形,
∴△ABD∽△A'B'D' .
∴
证明:如图,分别作△ABC和△A'B'C'的对应高AD和A'D'.
所以相似三角形对应高的比等于相似比.
探究新知
D′
C′
B′
A′
学生思考、... text has been truncated due to evaluation version limitation.
8
(2)如图,分别作△ABC和△A'B'C'的对应角平分线AD和A'D'.
A′
C′
B′
A
C
B
D
D′
探究新知
学生思考、... text has been truncated due to evaluation version limitation.
9
∴∠BAD= ∠BAC= ∠B'A'C'=∠B'A'D'.
∴在△ABD与△A'B'D'中,
△ABD∽△A'B'D'.
∴ .
∵△ABC∽△A'B'C',
∴∠B=∠B',∠BAC=∠B'A'C'.
∵AD和A'D'分别是∠BAC和∠B'A'C'的平分线,
探究新知
学生思考、... text has been truncated due to evaluation version limitation.
10
(3)如图,分别作△ABC和△A'B'C'的对应中线AD和A'D'.
A′
C′
B′
A
C
B
D
D′
探究新知
学生思考、... text has been truncated due to evaluation version limitation.
11
∵△ABC∽△A'B'C',
∴∠B=∠B',