题目（八年级 · 画法几何学）：如图,在正方形$ABCD$中,点$G$在对角线$BD$上(不与点$B,D$重合),$GEperpDC$于点$E,GF$$perpBC$于点$F$,连接$AG$.(1)写出线段$AG,GE,GF$长度之间的数量关系,并说明理由;(2)若正方形$ABCD$的边长为$1,angleAGF=105^{circ}$,求线段$BG$的长.
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答案：(1)结论:$AG^{2}=GE^{2}+GF^{2}$.只要证明$GA=GC$,四边形$EGFC$是矩形,推出$GE=CF$,在Rt$triangleGFC$中，利用勾股定理即可证明;(2)过点$A$作$AHperpBG$,在Rt$triangleABH、$Rt$triangleAHG$中,求出$AH、HG$即可解决问题.【解析】解:(1)结论:$AG^{2}=GE^{2}+GF^{2}$.理由:连接$CG$.$because$四边形$ABCD$是正方形,$thereforeA、C$关于对角线$BD$对称，$because$点$G$在$BD$上,$thereforeGA=GC$,$becauseGEperpDC$于点$E,GFperpBC$于点$F$,$thereforeangleGEC=angleECF=angleCFG=90^{circ}$,$therefore$四边形$EGFC$是矩形,$thereforeCF=GE$,在Rt$triangleGFC$中,$becauseCG^{2}=GF^{2}+CF^{2}$,$thereforeAG^{2}=GF^{2}+GE^{2}$.(2)过点$A$作$AHperpBG$,$because$四边形$ABCD$是正方形,$thereforeangleABD=angleGBF=45^{circ}$,$becauseGFperpBC$,$thereforeangleBGF=45^{circ}$,$becauseangleAGF=105^{circ}$,$thereforeangleAGB=angleAGF-angleBGF=105^{circ}-45^{circ}=60^{circ}$,在Rt$triangleABH$中,$becauseAB=1$,$thereforeAH=BH=frac{sqrt{2}}{2}$,在Rt$triangleAGH$中,$becauseAH=frac{sqrt{2}}{2},angleGAH=30^{circ}$,$thereforeHG=AHcdottan30^{circ}=frac{sqrt{6}}{6}$,$thereforeBG=BH+HG=frac{sqrt{2}}{2}+frac{sqrt{6}}{6}$.解法二:如图,过点$B$作$BNperpAG$于$N$,在$BN$上截取$BM$,使得$BM=AM$,设$AN=x$.<ImageHere>$becauseangleAGF=105^{circ},angleGBF=angleFGB=angleABG=45^{circ}$,$thereforeangleAGB=60^{circ},angleGBN=30^{circ},angleABM=angleBAM=15^{circ}$,$thereforeangleAMN=angleABM+angleBAM=30^{circ}$,$thereforeAM=BM=2x,quadMN=sqrt{3}x$,在Rt$triangleABN$中,则有$1=x^{2}+(2x+sqrt{3}x)^{2}$,解得$x=frac{sqrt{6}-sqrt{2}}{4}$,$thereforeBN=frac{sqrt{6}-sqrt{2}}{4}$,$thereforeBG=frac{BN}{cos30^{circ}}=frac{3sqrt{2}+sqrt{6}}{6}$.<ImageHere>【点睛】本题考查正方形的性质、矩形的判定和性质、勾股定理直角三角形30度的性质等知识,解题的关键是学会添加常用辅助线,构造直角三角形解决问题,属于中考常考题型.
解析：(1)结论:$AG^{2}=GE^{2}+GF^{2}$.只要证明$GA=GC$,四边形$EGFC$是矩形,推出$GE=CF$,在Rt$triangleGFC$中，利用勾股定理即可证明;(2)过点$A$作$AHperpBG$,在Rt$triangleABH、$Rt$triangleAHG$中,求出$AH、HG$即可解决问题.【解析】解:(1)结论:$AG^{2}=GE^{2}+GF^{2}$.理由:连接$CG$.$because$四边形$ABCD$是正方形,$thereforeA、C$关于对角线$BD$对称，$because$点$G$在$BD$上,$thereforeGA=GC$,$becauseGEperpDC$于点$E,GFperpBC$于点$F$,$thereforeangleGEC=angleECF=angleCFG=90^{circ}$,$therefore$四边形$EGFC$是矩形,$thereforeCF=GE$,在Rt$triangleGFC$中,$becauseCG^{2}=GF^{2}+CF^{2}$,$thereforeAG^{2}=GF^{2}+GE^{2}$.(2)过点$A$作$AHperpBG$,$because$四边形$ABCD$是正方形,$thereforeangleABD=angleGBF=45^{circ}$,$becauseGFperpBC$,$thereforeangleBGF=45^{circ}$,$becauseangleAGF=105^{circ}$,$thereforeangleAGB=angleAGF-angleBGF=105^{circ}-45^{circ}=60^{circ}$,在Rt$triangleABH$中,$becauseAB=1$,$thereforeAH=BH=frac{sqrt{2}}{2}$,在Rt$triangleAGH$中,$becauseAH=frac{sqrt{2}}{2},angleGAH=30^{circ}$,$thereforeHG=AHcdottan30^{circ}=frac{sqrt{6}}{6}$,$thereforeBG=BH+HG=frac{sqrt{2}}{2}+frac{sqrt{6}}{6}$.解法二:如图,过点$B$作$BNperpAG$于$N$,在$BN$上截取$BM$,使得$BM=AM$,设$AN=x$.$becauseangleAGF=105^{circ},angleGBF=angleFGB=angleABG=45^{circ}$,$thereforeangleAGB=60^{circ},angleGBN=30^{circ},angleABM=angleBAM=15^{circ}$,$thereforeangleAMN=angleABM+angleBAM=30^{circ}$,$thereforeAM=BM=2x,quadMN=sqrt{3}x$,在Rt$triangleABN$中,则有$1=x^{2}+(2x+sqrt{3}x)^{2}$,解得$x=frac{sqrt{6}-sqrt{2}}{4}$,$thereforeBN=frac{sqrt{6}-sqrt{2}}{4}$,$thereforeBG=frac{BN}{cos30^{circ}}=frac{3sqrt{2}+sqrt{6}}{6}$.【点睛】本题考查正方形的性质、矩形的判定和性质、勾股定理直角三角形30度的性质等知识,解题的关键是学会添加常用辅助线,构造直角三角形解决问题,属于中考常考题型.