如图1,在△ABC中,D、E、F分别为三边的中点,G点在边AB上,且DG平分△ABC的周长,设BC=a、AC=b、AB=c.
如图1,在△ABC中,D、E、F分别为三边的中点,G点在边AB上,且DG平分△ABC的周长,设BC=a、AC=b、AB=c.(1)求线段BG的长;(2)求证:DG平分∠E...
如图1,在△ABC中,D、E、F分别为三边的中点,G点在边AB上,且DG平分△ABC的周长,设BC=a、AC=b、AB=c.(1)求线段BG的长;(2)求证:DG平分∠EDF;(3)连接CG,如图2,若△GBD ∽△GDF,求证:BG⊥CG.
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妄想与你丿06
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(1) ![](https://iknow-pic.cdn.bcebos.com/9358d109b3de9c829537ac3d6f81800a19d84335?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) (b+c);(2)证明见解析;(3)证明见解析. |
试题分析:(1)由△BDG与四边形ACDG的周长相等与BD=CD,易得BG=AC+AG,即可得BG= ![](https://iknow-pic.cdn.bcebos.com/9358d109b3de9c829537ac3d6f81800a19d84335?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) (AB+AC); (2)由点D、F分别是BC、AB的中点,利用三角形中位线的性质,易得DF= ![](https://iknow-pic.cdn.bcebos.com/9358d109b3de9c829537ac3d6f81800a19d84335?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) AC= ![](https://iknow-pic.cdn.bcebos.com/9358d109b3de9c829537ac3d6f81800a19d84335?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) b,由FG=BG-BF,求得DF=FG,又由DE∥AB,即可求得∠FDG=∠EDG; (3)由△BDG与△DFG相似,∠DFG>∠B,∠BGD=∠DGF(公共角),可得∠B=∠FDG,又由(2)得:∠FGD=∠FDG,易证得DG=BD=CD,可得B、G、C三点在以BC为直径的圆周上,由圆周角定理,即可得BG⊥CG. 试题解析:(1)解:∵△BDG与四边形ACDG的周长相等, ∴BD+BG+DG=AC+CD+DG+AG, ∵D是BC的中点, ∴BD=CD, ∴BG=AC+AG, ∵BG+(AC+AG)=AB+AC, ∴BG= ![](https://iknow-pic.cdn.bcebos.com/9358d109b3de9c829537ac3d6f81800a19d84335?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) (AB+AC)= ![](https://iknow-pic.cdn.bcebos.com/9358d109b3de9c829537ac3d6f81800a19d84335?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) (b+c); (2)证明:∵点D、F分别是BC、AB的中点, ∴DF= ![](https://iknow-pic.cdn.bcebos.com/9358d109b3de9c829537ac3d6f81800a19d84335?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) AC= ![](https://iknow-pic.cdn.bcebos.com/9358d109b3de9c829537ac3d6f81800a19d84335?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) b,BF= ![](https://iknow-pic.cdn.bcebos.com/9358d109b3de9c829537ac3d6f81800a19d84335?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) AB= ![](https://iknow-pic.cdn.bcebos.com/9358d109b3de9c829537ac3d6f81800a19d84335?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) c, 又∵FG=BG-BF= ![](https://iknow-pic.cdn.bcebos.com/9358d109b3de9c829537ac3d6f81800a19d84335?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) (b+c)- ![](https://iknow-pic.cdn.bcebos.com/9358d109b3de9c829537ac3d6f81800a19d84335?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) c= ![](https://iknow-pic.cdn.bcebos.com/9358d109b3de9c829537ac3d6f81800a19d84335?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) b, ∴DF=FG, ∴∠FDG=∠FGD, ∵点D、E分别是BC、AC的中点, ∴DE∥AB, ∴∠EDG=∠FGD, ∴∠FDG=∠EDG, 即DG平分∠EDF; (3)证明:∵△BDG与△DFG相似,∠DFG>∠B,∠BGD=∠DGF(公共角), ∴∠B=∠FDG, 由(2)得:∠FGD=∠FDG, ∴∠FGD=∠B, ∴DG=BD, ∵BD=CD, ∴DG=BD=CD, ∴B、G、C三点在以BC为直径的圆周上, ∴∠BGC=90°, 即BG⊥CG. |
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