(1)在△ABC中,∠ACB=90°,CD⊥AB于D,设AC=b,BC=a,AB=c,CD=h.求证:(1) 1 a 2

2024-11-24 00:03:43
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回答1:

(1)证明:∵∠ACB=90°,
∴a 2 +b 2 =c 2 ,S △ABC =
1
2
AC?BC=
1
2
ab.
∵CD⊥AB于D,
∴S △ABC =
1
2
AB?CD=
1
2
ch.
1
2
ab=
1
2
ch,
∴ab=ch,
c
ab
=
1
h

c 2
a 2 b 2
=
1
h 2

∵a 2 +b 2 =c 2
a 2 + b 2
a 2 b 2
=
1
h 2

a 2
a 2 b 2
+
b 2
a 2 b 2
=
1
h 2

1
a 2
+
1
b 2
=
1
h 2


(2)以a+b,h和c+h为边构成的三角形是直角三角形,
∵(a+b) 2 +h 2 =a 2 +2ab+b 2 +h 2 =c 2 +2ab+h 2 ,(c+h) 2 =c 2 +2ch+h 2
∵ab=ch,
∴(a+b) 2 +h 2 =(c+h) 2
∴以a+b,h和c+h为边构成的三角形是直角三角形.