正方形ABCD中,E,F分别在AD,DC上,∠ABE=∠CBF=15°,G是AD上另一点,且∠BGD=120°,连接EF,BG,FG,EF与BG交于点H.求证:BG=EG+FG
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![正方形ABCD中,E,F分别在AD,DC上,∠ABE=∠CBF=15°,G是AD上另一点,且∠BGD=120°,连接EF,BG,FG,EF与BG交于点H.求证:BG=EG+FG](/uploads/image/z/5618149-61-9.jpg?t=%E6%AD%A3%E6%96%B9%E5%BD%A2ABCD%E4%B8%AD%2CE%2CF%E5%88%86%E5%88%AB%E5%9C%A8AD%2CDC%E4%B8%8A%2C%E2%88%A0ABE%3D%E2%88%A0CBF%3D15%C2%B0%2CG%E6%98%AFAD%E4%B8%8A%E5%8F%A6%E4%B8%80%E7%82%B9%2C%E4%B8%94%E2%88%A0BGD%3D120%C2%B0%2C%E8%BF%9E%E6%8E%A5EF%2CBG%2CFG%2CEF%E4%B8%8EBG%E4%BA%A4%E4%BA%8E%E7%82%B9H.%E6%B1%82%E8%AF%81%EF%BC%9ABG%3DEG%2BFG)
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