初中数学几何题

\u521d\u4e2d\u6570\u5b66\u9898\u76ee\uff0c\u51e0\u4f55\u9898

\u3010\u9898\u76ee\u3011
\u5df2\u77e5\u5728\u25b3ABC\u4e2d\uff0c\u2220CAB=2\u03b1\uff0c\u4e140\uff1c\u03b1\uff1c30\u00b0\uff0cAP\u5e73\u5206\u2220CAB\uff0c\u82e5\u2220ABC=60\u00b0-\u03b1\uff0c\u70b9P\u5728\u25b3ABC\u7684\u5185\u90e8\uff0c\u4e14\u4f7f\u2220CBP=30\u00b0\uff0c\u6c42\u2220APC\u7684\u5ea6\u6570\uff08\u7528\u542b\u03b1\u7684\u4ee3\u6570\u5f0f\u8868\u793a\uff09\u3002

\u3010\u89e3\u7b54\u3011

\u3010\u89e3\u6cd5\u4e00\u3011
\u89e3\uff1a
\u5ef6\u957fAC\u81f3M\uff0c\u4f7fAM=AB\uff0c\u8fde\u63a5PM\uff0cBM\uff08\u5982\u56fe1\uff09
\u2235AP\u5e73\u5206\u2220CAB\uff0c\u2220CAB=2\u03b1
\u2234\u22201=\u22202= \u03b1
\u5728\u25b3AMP\u548c\u25b3ABP\u4e2d\uff1a
\u2235AM=AB\uff0c\u22201 =\u22202\uff0cAP=AP
\u2234\u25b3AMP\u224c\u25b3ABP
\u2234PM=PB\uff0c\u22203 =\u22204
\u2235\u2220ABC=60\u00b0\uff0d\u03b1\uff0c\u2220CBP=30\u00b0
\u2234\u22204=(60\u00b0\uff0d\u03b1)\uff0d30\u00b0=30\u00b0\uff0d\u03b1
\u2234\u22203 =\u22204 =30\u00b0\uff0d\u03b1
\u2235\u25b3AMB\u4e2d\uff0cAM=AB
\u2234\u2220AMB=\u2220ABM=(180\u00b0\uff0d\u2220MAB)\u00f72 =(180\u00b0\uff0d2\u03b1)\u00f72 =90\u00b0\uff0d\u03b1
\u2234\u22205=\u2220AMB\uff0d\u22203= (90\u00b0\uff0d\u03b1)\uff0d(30\u00b0\uff0d\u03b1)=60\u00b0
\u2234\u25b3PMB\u4e3a\u7b49\u8fb9\u25b3
\u2235\u22206=\u2220ABM\uff0d\u2220ABC = (90\u00b0\uff0d\u03b1)\uff0d(60\u00b0\uff0d\u03b1)=30\u00b0
\u2234\u22206=\u2220CBP
\u2234BC\u5e73\u5206\u2220PBM
\u2234BC\u5782\u76f4\u5e73\u5206PM
\u2234CP=CM
\u2234\u22207 =\u22203 = 30\u00b0\uff0d\u03b1
\u2234\u2220ACP=\u22207\uff0b\u22203=(30\u00b0\uff0d\u03b1)\uff0b(30\u00b0\uff0d\u03b1)=60\u00b0\uff0d2\u03b1
\u2234\u25b3ACP\u4e2d\uff0c\u2220APC=180\u00b0\uff0d\u22201\uff0d\u2220ACP
=180\u00b0\uff0d\u03b1\uff0d(60\u00b0\uff0d2\u03b1)
=120\u00b0\uff0b\u03b1


\u3010\u89e3\u6cd5\u4e8c\u3011
\u89e3\uff1a
\u5728AB\u4e0a\u622a\u53d6AM\uff0c\u4f7fAM=AC\uff0c\u8fde\u63a5PM\uff0c\u5ef6\u957fAP\u4ea4BC\u4e8eN\uff0c\u8fde\u63a5MN\uff08\u5982\u56fe2\uff09
\u2235AP\u5e73\u5206\u2220CAB\uff0c\u2220CAB=2\u03b1
\u2234\u22201=\u22202=\u03b1
\u5728\u25b3ACN\u548c\u25b3AMN\u4e2d\uff1a
\u2235AC=AM\uff0c\u22201 =\u22202\uff0c AN=AN
\u2234\u25b3ACN\u224c\u25b3AMN
\u2234\u22203 =\u22204
\u2235\u2220ABC=60\u00b0\uff0d\u03b1
\u2234\u22203=\u22202\uff0b\u2220NBA=\u03b1\uff0b(60\u00b0\uff0d\u03b1) =60\u00b0
\u2234\u22203 =\u22204 =60\u00b0
\u2234\u22205=180\u00b0\uff0d\u22203\uff0d\u22204=180\u00b0\uff0d60\u00b0\uff0d60\u00b0=60\u00b0
\u2234\u22204 =\u22205
\u2234NM\u5e73\u5206\u2220PNB
\u2235\u2220CBP=30\u00b0
\u2234\u22206=\u22203\uff0d\u2220NBP=60\u00b0\uff0d30\u00b0=30\u00b0
\u2234\u22206=\u2220NBP
\u2234NP=NB
\u2234NM\u5782\u76f4\u5e73\u5206PB
\u2234MP=MB
\u2234\u22207 =\u22208
\u2234\u22206\uff0b\u22207 =\u2220NBP\uff0b\u22208
\u5373\u2220NPM=\u2220NBM =60\u00b0\uff0d\u03b1
\u2234\u2220APM=180\u00b0\uff0d\u2220NPM =180\u00b0\uff0d(60\u00b0\uff0d\u03b1)=120\u00b0\uff0b\u03b1
\u5728\u25b3ACP\u548c\u25b3AMP\u4e2d\uff1a
\u2235AC=AM\uff0c \u22201 =\u22202\uff0c AP=AP
\u2234\u25b3ACP\u224c\u25b3AMP
\u2234\u2220APC=\u2220APM
\u2234\u2220APC=120\u00b0\uff0b\u03b1

\u8fd9\u91cc\u6709\u51e0\u4f55\u7684\u89e3\u9898\u65b9\u6cd5\u4e5f\u8bb8\u6709\u7528

解：分为两种情况求解：
（1）AB=AD+BE+DE：
此时，容易证明△AEC∽△BCD
AE/BC=AC/BD
即：AC*BC=AC^2=(AB-4)*(AB-3)=AB^2-7AB+12
AB^2=2*AC^2
解得AB=12，DE=5
S△ABC=36

（2）AB=AE+BD-DE
此时，容易证明△ADC∽△BCE
AD/BC=AC/BE
AC*BC=AC^2=AD*BE=12,
AC=2根号3
S△ABC=6

AD=3 BE=4 DE=7 AB=14 AC=BC=7根号2

  面积ABC=14

旋转ECB使AC与BC重合得三角形FCA 因为角FCD=角DCE CB=CF DC=CD所以CDF全等DCB 所以FD=DE=FA+AD=3+4=7 所以AE=FE=7+7=14 而AC=AB 所以由勾股定理得AC=BC=根号98 所以ABC的面积=49

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