题目内容
如图所示,在△ABC中,∠BAC=135°,EF、GH分别是AB、AC两边的垂直平分线,与BC边交于点E、G,求∠EAG的度数.
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∵EF、GH分别是AB、AC两边的垂直平分线,
∴AE=BE,AG=CG,
∴∠BAE=∠B,∠CAG=∠C,
∵∠BAC=135°,
∴∠B+∠C=180°-∠BAC=45°,
∴∠BAE+∠CAG=45°,
∴∠EAG=∠BAC-(∠BAE+∠CAG)=135°-45°=90°.

∴AE=BE,AG=CG,
∴∠BAE=∠B,∠CAG=∠C,
∵∠BAC=135°,
∴∠B+∠C=180°-∠BAC=45°,
∴∠BAE+∠CAG=45°,
∴∠EAG=∠BAC-(∠BAE+∠CAG)=135°-45°=90°.
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