如图6,在由小正方形组成的网格中,小正方形的顶点称为格点,已知点A,B,C都在格点上。
\u5982\u56fe\uff0c\u5728\u8fb9\u957f\u4e3a1\u7684\u5c0f\u6b63\u65b9\u5f62\u7ec4\u6210\u7684\u7f51\u683c\u4e2d\uff0c\u25b3AOB\u7684\u9876\u70b9\u90fd\u5728\u683c\u70b9\u4e0a\uff0c\u70b9A\u3001B\u7684\u5750\u6807\u5206\u522b\u4e3a\uff08\uff0d4\uff0c4\uff09\u3001\uff08\uff0d6\uff081\uff09A 1 (\uff0d4\uff0c8)\u3001B 1 (\uff0d6\uff0c6) \uff082\uff098\u03c0 \u8bd5\u9898\u5206\u6790\uff1a\uff0e\u2474 \u6839\u636e\u5e73\u79fb\u6cd5\u5219\uff0c\u5411\u4e0a\u5e73\u79fb\uff0c\u5219x\u8f74\u5750\u6807\u4e0d\u53d8\uff0cy\u8f74\u589e\u52a04\u4e2a\u5355\u4f4d\uff0c\u5373A 1 (\uff0d4\uff0c8)\u3001B 1 (\uff0d6\uff0c6) \u2475 \u5982\u56fe\uff1a \u65cb\u8f6c90\u5ea6\uff0c\u5219\u7ebf\u6bb5AO\u6240\u626b\u8fc7\u7684\u9762\u79ef\u662f\u6247\u5f62\uff0c\u5706\u5468\u89d2\u4e3a90\u5ea6\uff0c\u534a\u5f84\u6839\u636e\u52fe\u80a1\u5b9a\u7406\u5f97\u51fa\u534a\u5f84OA= \uff0c\u6240\u4ee5\u6839\u636e\u9762\u79ef\u516c\u5f0f\u5f97\u51fa =8\u03c0\uff0e \u2476 \u5982\u56fe \u70b9\u8bc4\uff1a\u96be\u5ea6\u7cfb\u6570\u4e2d\u7b49\u3002\u8003\u67e5\u5e73\u9762\u56fe\u5f62\u7684\u5e73\u79fb\u548c\u65cb\u8f6c\uff0c\u4e00\u6761\u7ebf\u6bb5\u65cb\u8f6c90\u5ea6\u5f97\u5230\u7684\u56fe\u5f62\u7684\u5f62\u72b6\u662f\u6247\u5f62\u3002
\u89e3\uff1a\uff081\uff09\u5982\u56fe\uff0c\u8fde\u63a5AC\uff0c\u7531\u52fe\u80a1\u5b9a\u7406\u5f97\uff0cAB2=12+22=5\uff0cBC2=12+22=5\uff0cAC2=12+32=10\uff0c\u2234AB2+BC2=AC2\uff0c\u2234\u25b3ABC\u662f\u76f4\u89d2\u4e09\u89d2\u5f62\uff0c\u2220ABC=90\u00b0\uff0c\u2234AB\u22a5BC\uff1b\uff082\uff09\u2220\u03b1+\u2220\u03b2=45\u00b0\uff0e\u8bc1\u660e\u5982\u4e0b\uff1a\u5982\u56fe\uff0c\u7531\u52fe\u80a1\u5b9a\u7406\u5f97\uff0cAB2=12+22=5\uff0cBC2=12+22=5\uff0cAC2=12+32=10\uff0c\u2234AB2+BC2=AC2\uff0c\u2234\u25b3ABC\u662f\u76f4\u89d2\u4e09\u89d2\u5f62\uff0c\u2235AB=BC\uff0c\u2234\u25b3ABC\u662f\u7b49\u8170\u76f4\u89d2\u4e09\u89d2\u5f62\uff0e
连结AD，∵BE=CF=2，∠BED=∠CFA=90°，DE=AF，
∴△BDE≌△CAF（SAS）
∴∠ACB=∠DBE，
由勾股定理得AD²=5，BD²=5，AB²=10,
∴AD²+BD²=AB²，AD=BD
∴△ABD是等腰RT△，
∴∠ABD=45°,
∴∠ACB+∠ABC=∠DBE+∠ABC=45°
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