如图,在Rt△ABC中,∠ACB=90°,CD⊥AB于D,设AC=b,BC=a,CD=h,AB=c,下面有3个命题:(1) 1

2024-11-23 10:20:20
推荐回答(2个)
回答1:

(1)∵Rt△ABC中,∠ACB=90°,CD⊥AB于D,AC=b,BC=a,CD=h,AB=c,
∴c=
a 2 + b 2

∴S △ABC =
1
2
ab=
1
2
ch,
∴h=
ab
c
,h 2 =
a 2 b 2
c 2

1
h 2
=
c 2
a 2 b 2
,即
1
h 2
=
a 2 + b 2
a 2 b 2
=
1
a 2
+
1
b 2
,故(1)正确;

(2)∵
1
2
ab=
1
2
ch,
∴ab=ch,即a 2 b 2 =c 2 h 2
∴(a+b) 2 -a 2 -b 2 =(c+h) 2 -c 2 -h 2
∴(c+h) 2 -(a+b) 2 =c 2 -a 2 -b 2 +h 2
∵a 2 +b 2 =c 2
∴(c+h) 2 -(a+b) 2 =h 2
∵h>0,且a b c h均为线段.
∴a>0,b>0,c>0,h>0,
∴c+h>a+b,故(3)正确;

(3)∵(c+h) 2 =c 2 +2ch+h 2
h 2 +(a+b) 2 =h 2 +a 2 +2ab+b 2 ,a 2 +b 2 =c 2 (勾股定理),ab=ch(面积公式推导),
∴c 2 +2ch+h 2 =h 2 +a 2 +2ab+b 2
∴(c+h) 2 =h 2 +(a+b) 2
∴根据勾股定理的逆定理知道以h,c+h,a+b为边构成的三角形是直角三角形,故正确.
故选D.

回答2:

哈哈,自己写