{"common":{"save":0,"post_id":"1357","level":3,"total":10,"point":10,"point_extra":0},"segment":[{"id":"1941","post_id":"1357","mon_id":"0","chapter_id":"0","question":"\u0110i\u1ec1n d\u1ea5u th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","options":{"time":24,"part":[{"title":"\u0110i\u1ec1n d\u1ea5u th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["120"]]],"list":[{"point":10,"width":50,"content":"","type_input":"","ques":"Cho tam gi\u00e1c $ABC$, tia ph\u00e2n gi\u00e1c g\u00f3c $A$ c\u1eaft c\u1ea1nh $BC$ \u1edf $D$. \u0110\u01b0\u1eddng th\u1eb3ng qua $D$ song song v\u1edbi $AB$ c\u1eaft $AC$ \u1edf $M$. T\u00ednh $\\widehat{AMD}$ bi\u1ebft $\\widehat{A} = 60^{o}$. <br\/> <b> \u0110\u00e1p \u00e1n l\u00e0: <\/b> $\\widehat{AMD}=$ _input_ $^{o}$","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> Ch\u1ee9ng minh $\\triangle{AMD}$ c\u00e2n t\u1ea1i $M$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> T\u00ednh $\\widehat{AMD}$ <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='img\/H7C3B21_K10.png' \/><\/center> <br\/> $\\blacktriangleright$ V\u00ec $AD$ l\u00e0 tia ph\u00e2n gi\u00e1c $\\widehat{A}$ n\u00ean $\\widehat{A_{1}} = \\widehat{A_{2}}$ (\u0111\u1ecbnh ngh\u0129a) (1) <br\/> M\u1eb7t kh\u00e1c $DM \/\/ AB$ (gt) n\u00ean $\\widehat{A_{1}} = \\widehat{D_{1}}$ (so le trong) (2) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow$ $\\widehat{A_{2}} = \\widehat{D_{1}}$ <br\/> $\\triangle{AMD}$ c\u00f3 $\\widehat{A_{2}} = \\widehat{D_{1}}$ n\u00ean $\\triangle{AMD}$ c\u00e2n t\u1ea1i $M$ <br\/> Ta c\u00f3: $\\widehat{AMD} + \\widehat{A_{2}} + \\widehat{D_{1}} = 180^{o}$ (t\u1ed5ng ba g\u00f3c trong $\\triangle{AMD}$) <br\/> $ \\begin{align} \\Rightarrow \\widehat{AMD} &= 180^{o} - (\\widehat{A_{2}} + \\widehat{D_{1}}) \\\\ &= 180^{o} - (\\widehat{A_{2}} + \\widehat{A_{2}}) (\\text{v\u00ec} \\hspace{0,2cm} \\widehat{A_{2}} = \\widehat{D_{1}}) \\\\ &= 180^{o} - \\widehat{A} (\\text{v\u00ec} \\hspace{0,2cm} \\widehat{A_{1}} = \\widehat{A_{2}}) \\\\ &= 180^{o} - 60^{o} \\\\ &= 120^{o} \\end{align}$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p c\u1ea7n \u0111i\u1ec1n l\u00e0: $120$ <\/span>"}]}]},"correct":"","level":"3","hint":"","answer":"","type":"json","extra_type":"","time":"0","user_id":"0","test":"0","date":"2019-09-30 09:23:06"},{"id":"1942","post_id":"1357","mon_id":"0","chapter_id":"0","question":"\u0110i\u1ec1n d\u1ea5u th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","options":{"time":24,"part":[{"title":"\u0110i\u1ec1n d\u1ea5u th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["EF","FE"]]],"list":[{"point":10,"width":50,"content":"","type_input":"","ques":"Cho tam gi\u00e1c $ABC$, \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $A$ v\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $B$ c\u1eaft nhau t\u1ea1i $O$. Qua $O$ k\u1ebb $EF \/\/ BC$ ($E \\in AB; F \\in AC$) <br\/> Khi \u0111\u00f3: $BE + CF =$ _input_ ","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> Ch\u1ee9ng minh $BE = EO$ b\u1eb1ng c\u00e1ch ch\u1ee9ng minh $\\triangle{BEO}$ c\u00e2n t\u1ea1i $E$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> Ch\u1ee9ng minh $OF = FC$ b\u1eb1ng c\u00e1ch ch\u1ee9ng minh $\\triangle{OFC}$ c\u00e2n t\u1ea1i $F$ <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='img\/H7C3B21_K05.png' \/><\/center> $\\blacktriangleright$ X\u00e9t $\\triangle{BOE}$ c\u00f3: <br\/> $\\widehat{B_{1}} = \\widehat{B_{2}}$ (gi\u1ea3 thi\u1ebft) <br\/> $\\widehat{O_{1}} = \\widehat{B_{2}}$ (so le trong) <br\/> $\\Rightarrow$ $\\widehat{B_{1}} = \\widehat{O_{1}}$ $\\Rightarrow$ $\\triangle{BEO}$ c\u00e2n t\u1ea1i $E$ <br\/> $\\Rightarrow$ $BE = EO$ (\u0111\u1ecbnh ngh\u0129a tam gi\u00e1c c\u00e2n) (1)<br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{CFO}$ c\u00f3: <br\/> $\\widehat{C_{1}} = \\widehat{C_{2}}$ (gi\u1ea3 thi\u1ebft) <br\/> $\\widehat{O_{2}} = \\widehat{C_{2}}$ (so le trong) <br\/> $\\Rightarrow$ $\\widehat{O_{2}} = \\widehat{C_{1}}$ $\\Rightarrow$ $\\triangle{OFC}$ c\u00e2n t\u1ea1i $F$ (t\u00ednh ch\u1ea5t) <br\/> $\\Rightarrow$ $OF = FC$ (\u0111\u1ecbnh ngh\u0129a tam gi\u00e1c c\u00e2n) (2) <br\/> T\u1eeb (1) v\u00e0 (2)$\\Rightarrow$ $BE + FC = EF$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p c\u1ea7n \u0111i\u1ec1n l\u00e0: $EF$ <\/span>"}]}]},"correct":"","level":"3","hint":"","answer":"","type":"json","extra_type":"","time":"0","user_id":"0","test":"0","date":"2019-09-30 09:23:06"},{"id":"1943","post_id":"1357","mon_id":"0","chapter_id":"0","question":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","options":{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"Cho tam gi\u00e1c $ABC$ c\u00f3 $\\widehat{A} = 90^{o}$. Tr\u00ean c\u1ea1nh $AC$ l\u1ea5y \u0111i\u1ec3m $D$ sao cho $\\widehat{ABC} = 3\\widehat{ABD}$. Tr\u00ean c\u1ea1nh $AB$ l\u1ea5y \u0111i\u1ec3m $E$ sao cho $\\widehat{ACB} = 3 \\widehat{ACE}$. G\u1ecdi $F$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $DB$ v\u00e0 $CE$, $I$ l\u00e0 giao \u0111i\u1ec3m c\u00e1c tia ph\u00e2n gi\u00e1c c\u1ee7a $\\triangle{BFC}$. Trong c\u00e1c kh\u1eb3ng \u0111\u1ecbnh sau kh\u1eb3ng \u0111\u1ecbnh n\u00e0o \u0111\u00fang? ","select":["A. $\\widehat{BFC} = 90^{o} $ ","B. $\\widehat{BFC} = 120^{o}$ ","C. $\\triangle{DEI}$ l\u00e0 tam gi\u00e1c \u0111\u1ec1u ","D. B v\u00e0 C \u0111\u00fang "],"hint":"","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> T\u00ednh $\\widehat{BFC}$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> So s\u00e1nh $EI; ED; DI$ \u0111\u1ec3 bi\u1ebft $\\triangle{ABC}$ \u0111\u1ec1u l\u00e0 \u0111\u00fang hay sai <br\/><br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='img\/H7C3B21_K01.png' \/><\/center> <br\/> Ta c\u00f3: <br\/> $\\blacktriangleright$ $\\widehat{B} + \\widehat{C} = 90^{o}$ (v\u00ec $\\triangle{ABC}$ vu\u00f4ng t\u1ea1i $A$) <br\/> V\u00ec $\\widehat{ABC} = 3\\widehat{ABD}$; $\\widehat{ACB} = 3\\widehat{ACE}$ (gt) <br\/> $\\Rightarrow$ $\\widehat{FBC} + \\widehat{FCB} = \\dfrac{2}{3}(\\widehat{B} + \\widehat{C}) = \\dfrac{2}{3}. 90^{o} = 60^{o}$ <br\/> Trong $\\triangle{FBC}$ c\u00f3: $\\widehat{FCB} + \\widehat{FBC} + \\widehat{BFC} = 180^{o}$ (t\u1ed5ng ba g\u00f3c trong $1$ tam gi\u00e1c) <br\/> $ \\begin{align} \\Rightarrow \\widehat{BFC} &= 180^{o} - (\\widehat{FBC} + \\widehat{FCB}) \\\\ &= 180^{o} - 60^{o} \\\\ &= 120^{o}\\end{align} $ <br\/> $\\blacktriangleright$ V\u00ec $\\widehat{BFC} = 120^{o}$ $\\Rightarrow$ $\\widehat{BFE} = \\widehat{CFD} = 180^{o} - 120^{o} = 60^{o}$ (c\u00f9ng k\u1ec1 b\u00f9 v\u1edbi $\\widehat{BFC}$) <br\/> V\u00ec $I$ l\u00e0 giao \u0111i\u1ec3m $3$ tia ph\u00e2n gi\u00e1c c\u1ee7a $\\triangle{BFC}$ <br\/> $\\Rightarrow$ $FI$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{BFC}$ <br\/> Do \u0111\u00f3 $\\widehat{BFI} = \\widehat{CFI} = \\dfrac{\\widehat{BFC}}{2} = \\dfrac{120^{o}}{2} = 60^{o}$ <br\/> $\\triangle{BFE} = \\triangle{BFI}$ (g.c.g) v\u00ec: <br\/> $\\begin{cases} \\widehat{BFE} = \\widehat{BFI} = 60^{o} (cmt) \\\\ BF \\hspace{0,2cm} \\text{chung} \\\\ \\widehat{EBF} = \\widehat{FBI} (\\widehat{ABD} = \\dfrac{1}{3}\\widehat{ABC}; \\widehat{DBI} = \\dfrac{1}{2}\\widehat{DBC}) \\end{cases}$ <br\/> $\\Rightarrow$ $BI = BE$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) <br\/> $\\blacktriangleright$ $\\triangle{BED} = \\triangle{BFI}$ (c.g.c) v\u00ec: <br\/> $\\begin{cases} BE = BI (cmt) \\\\ \\widehat{EBD} = \\widehat{IBD} (\\triangle{EBF} = \\triangle{IBF}) \\\\ BD \\hspace{0,2cm} \\text{chung} \\end{cases}$ <br\/> $\\Rightarrow$ $ED = DI$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) (1) <br\/> $\\blacktriangleright$ $\\triangle{CFD} = \\triangle{CFI}$ (g.c.g) v\u00ec: <br\/> $\\begin{cases} \\widehat{CFD} = \\widehat{CFI} = 60^{o} (cmt) \\\\ FC \\hspace{0,2cm} \\text{chung} \\\\ \\widehat{ICF} = \\widehat{DCF} (\\widehat{ACB} = 3\\widehat{ACE}; \\widehat{ECI} = \\widehat{BCI}) \\end{cases}$ <br\/> $\\Rightarrow$ $CI = CD$ <br\/> $\\triangle{CDE} = \\triangle{CIE}$ (c.g.c) v\u00ec: <br\/> $\\begin{cases} CD = CI (cmt) \\\\ \\widehat{ICF} = \\widehat{DCF} (\\triangle{ICF} = \\triangle{DCF}) \\\\ FC \\hspace{0,2cm} \\text{chung} \\end{cases}$ <br\/> $\\Rightarrow$ $DE = EI$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) (2) <br\/> T\u1eeb (1) v\u00e0 (2) suy ra $DE = EI = ID$ <br\/> $\\Rightarrow$ $\\triangle{EID}$ l\u00e0 tam gi\u00e1c \u0111\u1ec1u <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: D <\/span>","column":2}]}]},"correct":"","level":"3","hint":"","answer":"","type":"json","extra_type":"","time":"0","user_id":"0","test":"0","date":"2019-09-30 09:23:06"}]}