Question

中考数学综合题几何。

已知Rt△ABC，∠C=90°，将△ABC逆时针旋转一定的角度，得到△A'BC',直线AA'与直线CC'交于点P，如图，当∠ABC=60°时，求证:AP=CC'。当∠ABC=45°时，线段AP和CC’的关系。

Answer 1

1)证明:作A'D⊥C'P于D,AE⊥C'P的延长线于E.

∵BC=BC'.

∴∠BC'C=∠BCC';又∠A'C'B=∠ACB=90°.

∴∠A'C'D=∠ACE(等角的余角相等)

又∵A'C'=AC;∠A'DC'=∠AEC=90°.

∴⊿A'DC'≌⊿AEC(AAS),A'D=AE

所以A'DP≌AEP容易证P是AA'的中点

因为30度比例是CB/BA=1/2

         C'BC相似于A'BA所以C'C=1/2A'A=AP

第2题同理啦，CB/BA=1/根2

            C'C=1/根2A'A=根2AP

Answer 2

，当∠ABC=60°时
设转动角度为a BC=1 AC=根号3
CC'=2sin(a/2) 角ACP=a/2 角APC=120
过A作PC延长线的垂线，垂足为H
所以AH=根号3 乘sina/2 所以APH=60 所以 AP=2sina/2

当∠ABC=45°时，
BC=1=AC 设转动角度为a
CC'=2sin(a/2) 角ACP=a/2 角APC=135
过A作PC延长线的垂线，垂足为H 所以APH=45
AH=sina/2 AP =根号2sina/2
所以CC'=根号2 AP

Answer 3

(1)证明:作A'D⊥C'P于D,AE⊥C'P的延长线于E.
∵BC=BC'.
∴∠BC'C=∠BCC';又∠A'C'B=∠ACB=90°.
∴∠A'C'D=∠ACE(等角的余角相等)
又∵A'C'=AC;∠A'DC'=∠AEC=90°.
∴⊿A'DC'≌⊿AEC(AAS),A'D=AE;
同理相似可证:⊿A'DP≌⊿AEP,得AP=A'P,即AA'=2AP;
∵∠ABC=∠A'BC'=60°.
∴∠ABA'=∠CBC';又AB=A'B,CB=C'B.
则⊿ABA'∽⊿CBC',得AA'/CC'=AB/BC=2/1,AA'=2CC'.
∴AP=CC'.
(2)解:当∠ABC=45°时,易知AC=BC,AB=√2BC,AB/BC=√2.
同理相似可证:AA'=2AP,AP=AA'/2;
⊿ABA'∽⊿CBC',AA'/CC'=AB/BC=√2, AA'=√2CC';
∴AP=AA'/2=(√2/2)CC'.