{"segment":[{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["90"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"Cho tam gi\u00e1c $ABC$ c\u00f3 $BC = 10cm$, c\u00e1c \u0111\u01b0\u1eddng trung tuy\u1ebfn $BD = 9cm$, trung tuy\u1ebfn $CE = 12cm$, $BD$ v\u00e0 $CE$ c\u1eaft nhau t\u1ea1i $G$. T\u00ednh g\u00f3c $\\widehat{BGC}$. <br\/> <b> \u0110\u00e1p \u00e1n l\u00e0: <\/b>$\\widehat{BGC}$ = _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> T\u00ednh $BG; CG$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> Ch\u1ee9ng minh $\\triangle{BGC}$ vu\u00f4ng t\u1ea1i $G$ <br\/> <b> B\u01b0\u1edbc 3: <\/b> T\u00ednh $\\widehat{BGC}$ <br\/><br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai20/lv3/img\/H7C3B20_K8.png' \/><\/center> <br\/> $\\blacktriangleright$ G\u1ecdi $G$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $BD$ v\u00e0 $CE$, suy ra $G$ l\u00e0 tr\u1ecdng t\u00e2m $\\triangle{ABC}$ <br\/> Theo t\u00ednh ch\u1ea5t tr\u1ecdng t\u00e2m, ta c\u00f3: <br\/> $BG = \\dfrac{2}{3}BD = \\dfrac{2}{3}.9 = 6(cm)$ <br\/> $CG = \\dfrac{2}{3}CE = \\dfrac{2}{3}.12 = 8(cm)$ <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{BGC}$ c\u00f3: $BC = 10cm; BG = 6cm; CG = 8cm$ <br\/> Ta c\u00f3: $10^2 = 6^2 + 8^2$ hay $BC^2 = CG^2 + BG^2$ <br\/> $\\Rightarrow$ $\\triangle{BGC}$ vu\u00f4ng t\u1ea1i $G$ <br\/> $\\Rightarrow$ $\\widehat{BGC} = 90^{o}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n l\u00e0: $90$ <\/span> "}]}],"id_ques":1911},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"Cho tam gi\u00e1c $ABC$ v\u1edbi trung tuy\u1ebfn $AD$. Qua $D$ k\u1ebb \u0111\u01b0\u1eddng th\u1eb3ng song song v\u1edbi $AB$, qua $B$ k\u1ebb \u0111\u01b0\u1eddng th\u1eb3ng song song v\u1edbi $AD$. Hai \u0111\u01b0\u1eddng th\u1eb3ng tr\u00ean c\u1eaft nhau t\u1ea1i $E$. Khi \u0111\u00f3 $D$ l\u00e0 tr\u1ecdng t\u00e2m tam gi\u00e1c $AEC$.<b> \u0110\u00fang <\/b> hay <b> sai <\/b>? ","select":["A. \u0110\u00daNG ","B. SAI "],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> G\u1ecdi $I$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $AE$ v\u00e0 $BC$ <br\/> Ch\u1ee9ng minh $DC = \\dfrac{2}{3}CI$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> Ch\u1ee9ng minh $CI$ l\u00e0 trung tuy\u1ebfn c\u1ee7a $\\triangle{ACE}$ <br\/> <b> B\u01b0\u1edbc 3 <\/b> : T\u1eeb b\u01b0\u1edbc 1 v\u00e0 b\u01b0\u1edbc 1 suy ra \u0111i\u1ec1u c\u1ea7n ch\u1ee9ng minh <br\/><br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai20/lv3/img\/H7C3B20_K9.png' \/><\/center> <br\/> $\\blacktriangleright$ G\u1ecdi $I$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $AE$ v\u00e0 $BC$ <br\/> V\u00ec $AB \/\/ DE$ v\u00e0 $AD \/\/ BE$ (gt) <br\/> $\\Rightarrow$ $AD = BE; AB = DE$ (t\u00ednh ch\u1ea5t \u0111o\u1ea1n ch\u1eafn song song) <br\/> V\u00e0 $\\widehat{BAI} = \\widehat{DEI}$ (so le trong) <br\/> $\\widehat{ABI} = \\widehat{EDI}$ (so le trong) <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{ABI}$ v\u00e0 $\\triangle{EDI}$ c\u00f3: <br\/> $\\begin{cases} \\widehat{ABI} = \\widehat{EDI} \\\\ AB = ED \\\\ \\widehat{BAI} = \\widehat{DEI} \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{ABI} = \\triangle{EDI}$ (g.c.g) <br\/> $\\Rightarrow$ $AI = EI$ v\u00e0 $IB = ID$ (c\u1eb7p c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) <br\/> V\u00ec $IB = ID(cmt)$ m\u00e0 $BD = DC$ $\\Rightarrow$ $DC = \\dfrac{2}{3}CI$ (1) <br\/> V\u00ec $AI = EI $ $\\Rightarrow$ $CI$ l\u00e0 trung tuy\u1ebfn c\u1ee7a $\\triangle{ACE}$ (2) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow$ $D$ l\u00e0 tr\u1ecdng t\u00e2m tam gi\u00e1c $ACE$ <br\/> V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 <span class='basic_pink'> \u0110\u00daNG <\/span>","column":2}]}],"id_ques":1912},{"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 $ABC$, \u0111\u01b0\u1eddng trung tuy\u1ebfn $AM$. Tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia $MA$ l\u1ea5y \u0111i\u1ec3m $D$ sao cho $MD = MA$. Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y \u0111\u00fang? ","select":["A. $AB \/\/ CD; AB = CD$ ","B. $AB > CD; AC < BD$","C. $AC \/\/ BD; AC = BD$ ","D. A 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> X\u00e9t $\\triangle{AMB}$ v\u00e0 $\\triangle{CMD}$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> X\u00e9t $\\triangle{AMC}$ v\u00e0 $\\triangle{BMD}$ <br\/> <b> B\u01b0\u1edbc 3: <\/b> D\u1ef1a v\u00e0o b\u01b0\u1edbc 1 v\u00e0 b\u01b0\u1edbc 2 \u0111\u1ec3 ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang. <br\/><br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai20/lv3/img\/H7C3B20_K3.png' \/><\/center> <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{AMB}$ v\u00e0 $\\triangle{CMD}$ c\u00f3: <br\/> $\\begin{cases} AM = MD (gt) \\\\ \\widehat{M_{3}} = \\widehat{M_{4}} (\\text{\u0111\u1ed1i} \\hspace{0,2cm} \\text{\u0111\u1ec9nh}) \\\\ BM = MC (gt) \\end{cases}$<br\/> $\\Rightarrow$ $\\triangle{AMB} = \\triangle{DMC}$ (c.g.c) <br\/> $\\Rightarrow$ $AB = CD$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) <br\/> $\\widehat{B_{1}} = \\widehat{C_{1}}$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) m\u00e0 hai g\u00f3c n\u00e0y \u1edf v\u1ecb tr\u00ed so le trong, b\u1eb1ng nhau n\u00ean $AB \/\/ CD$ <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{AMC}$ v\u00e0 $\\triangle{BMD}$ c\u00f3: <br\/> $\\begin{cases} AM = MD (gt) \\\\ MC = MB (gt) \\\\ \\widehat{M_{1}} = \\widehat{M_{2}} (\\text{\u0111\u1ed1i} \\hspace{0,2cm} \\text{\u0111\u1ec9nh}) \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{AMC} = \\triangle{DMB}$ (c.g.c) <br\/> $\\Rightarrow$ $AC = BD$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) <br\/> $\\widehat{A_{1}} = \\widehat{D_{1}}$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) m\u00e0 $\\widehat{A_{1}}$ v\u00e0 $\\widehat{D_{1}}$ \u1edf v\u1ecb tr\u00ed so le trong, b\u1eb1ng nhau <br\/> $\\Rightarrow$ $AC \/\/ BD$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: D <\/span>","column":2}]}],"id_ques":1913},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"Cho tam gi\u00e1c $ABC$, trung tuy\u1ebfn $AI$ v\u00e0 trung tuy\u1ebfn $BN$ c\u1eaft nhau \u1edf $O$. Tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia $IA$ l\u1ea5y \u0111i\u1ec3m $E$ sao cho $IE = IO$. Khi \u0111\u00f3 \u0111\u1ed9 d\u00e0i c\u00e1c c\u1ea1nh c\u1ee7a tam gi\u00e1c $BOE$ b\u1eb1ng $\\dfrac{2}{3}$ \u0111\u1ed9 d\u00e0i c\u00e1c \u0111\u01b0\u1eddng trung tuy\u1ebfn c\u1ee7a tam gi\u00e1c $ABC$. <b> \u0110\u00fang <\/b> hay <b> sai <\/b>? ","select":["A. \u0110\u00daNG ","B. SAI "],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> So s\u00e1nh $BO$ v\u1edbi $BN$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> So s\u00e1nh $OE = OI + IE$ v\u1edbi $AI$ <br\/> <b> B\u01b0\u1edbc 3: <\/b> Ch\u1ee9ng minh $BE = OC$ r\u1ed3i so s\u00e1nh v\u1edbi $CK$ <br\/><br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai20/lv3/img\/H7C3B20_K5.png' \/><\/center> <br\/> $\\blacktriangleright$ $O$ l\u00e0 tr\u1ecdng t\u00e2m tam gi\u00e1c $ABC$ (gt), suy ra $AI$ l\u00e0 trung tuy\u1ebfn <br\/> X\u00e9t $\\triangle{BOE}$ c\u00f3: $BO = \\dfrac{2}{3}BN$ (t\u00ednh ch\u1ea5t tr\u1ecdng t\u00e2m) (1) <br\/> $\\blacktriangleright$ Ta c\u00f3: $OI = \\dfrac{1}{3}AI$ (t\u00ednh ch\u1ea5t tr\u1ecdng t\u00e2m) <br\/> M\u00e0 $OI = IE$ (gt) $\\Rightarrow$ $OI = IE = \\dfrac{1}{3}AI$ <br\/> $\\Rightarrow$ $OE = OI + IE = \\dfrac{1}{3}AI + \\dfrac{1}{3}AI = \\dfrac{2}{3}AI$ (2) <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{BIE}$ v\u00e0 $\\triangle{CIO}$ c\u00f3: <br\/> $\\begin{cases} OI = IE (gt) \\\\ BI = IC (gt) \\\\ \\widehat{I_{1}} = \\widehat{I_{2}} (\\text{\u0111\u1ed1i} \\hspace{0.2cm} \\text{\u0111\u1ec9nh}) \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{BIE} = \\triangle{CIO}$ (c.g.c) <br\/> $\\Rightarrow$ $BE = CO$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) <br\/> M\u00e0 $CO = \\dfrac{2}{3}CK$ (t\u00ednh ch\u1ea5t tr\u1ecdng t\u00e2m) n\u00ean $BE = \\dfrac{2}{3}CK$ (3) <br\/> T\u1eeb (1), (2), (3) $\\Rightarrow$ $\\triangle{BOE}$ c\u00f3 ba c\u1ea1nh: <br\/> $OE = \\dfrac{2}{3}AI$; $BO = \\dfrac{2}{3}BN$; $BE = \\dfrac{2}{3}CK$ <br\/> V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 <span class='basic_pink'> \u0110\u00daNG <\/span>","column":2}]}],"id_ques":1914},{"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 $AB = c$; $AC = b; BC = a$ ($AB < AC$), c\u00e1c trung tuy\u1ebfn $AM, BN, CP$. <br\/> Trong c\u00e1c kh\u1eb3ng \u0111\u1ecbnh sau kh\u1eb3ng \u0111\u1ecbnh n\u00e0o \u0111\u00fang. ","select":["<span class='basic_left'> A. $BN + CP > \\dfrac{3a}{2}$ <\/span> "," <span class='basic_left'> B. $\\dfrac{1}{2}(b - c) < AM < \\dfrac{1}{2}(b + c)$ <\/span>"," <span class='basic_left'> C. $\\dfrac{3}{4}(a + b + c) < AM + BN + CP < a + b + c$ <\/span>"," <span class='basic_left'> D. C\u1ea3 3 \u0111\u00e1p \u00e1n tr\u00ean \u0111\u1ec1u \u0111\u00fang <\/span>"],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> X\u00e9t \u0111\u00e1p \u00e1n A <br\/> \u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c, so s\u00e1nh $GB + GC$ v\u1edbi $BC$, t\u1eeb \u0111\u00f3 so s\u00e1nh $BN + CP$ v\u1edbi $BC$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> X\u00e9t \u0111\u00e1p \u00e1n B <br\/> \u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c v\u00e0 h\u1ec7 qu\u1ea3 v\u1edbi tam gi\u00e1c $ABD$ \u0111\u1ec3 c\u00f3 \u0111i\u1ec1u c\u1ea7n t\u00ecm <br\/> <b> B\u01b0\u1edbc 3: <\/b> X\u00e9t \u0111\u00e1p \u00e1n C <br\/> \u00c1p d\u1ee5ng c\u00e1ch ch\u1ee9ng minh c\u00e2u $A$ v\u00e0 c\u00e2u $B$ \u0111\u1ec3 ch\u1ee9ng minh c\u00e2u $C$ <br\/><br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai20/lv3/img\/H7C3B20_K1.png' \/><\/center> <br\/> G\u1ecdi $G$ l\u00e0 tr\u1ecdng t\u00e2m c\u1ee7a tam gi\u00e1c $ABC$ <br\/> $\\blacktriangleright$ A - \u0110\u00fang v\u00ec: <br\/> X\u00e9t $\\triangle{ABC}$ c\u00f3: <br\/> $GB + GC > BC$ (b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c) <br\/> M\u1eb7t kh\u00e1c ta c\u00f3: $GB = \\dfrac{2}{3}BN; GC = \\dfrac{2}{3}CP$ (t\u00ednh ch\u1ea5t tr\u1ecdng t\u00e2m) <br\/> $\\Rightarrow$ $\\dfrac{2}{3}BN + \\dfrac{2}{3}CP > BC = a$ <br\/> $\\Leftrightarrow$ $BN + CP > \\dfrac{3a}{2}$ (1) <br\/> $\\blacktriangleright$ B - \u0110\u00fang v\u00ec: <br\/> Tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia $MA$ l\u1ea5y \u0111i\u1ec3m $D$ sao cho $MA = MD$ <br\/> X\u00e9t $\\triangle{AMC}$ v\u00e0 $\\triangle{DMB}$ c\u00f3: <br\/> $\\begin{cases} AM = MD (\\text{c\u00e1ch} \\hspace{0,2cm} \\text{l\u1ea5y} \\hspace{0,2cm} \\text{\u0111i\u1ec3m} D) \\\\ \\widehat{AMC} = \\widehat{BMB} (\\text{\u0111\u1ed1i} \\hspace{0,2cm} \\text{\u0111\u1ec9nh}) \\\\ BM = MC (gt) \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{AMC} = \\triangle{BMD}$ (c.g.c) <br\/> $\\Rightarrow$ $AC = BD$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) <br\/> $\\triangle{ABD}$ c\u00f3 $BD - AB < AD < BD + AB$ (b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c) <br\/> Hay $AC - AB < 2AM < AC + AB$ <br\/> $\\Rightarrow$ $\\dfrac{b - c}{2} < AM < \\dfrac{b + c}{2}$ <br\/> $\\Rightarrow$ $AM < \\dfrac{b + c}{2}$ (2) <br\/> $\\blacktriangleright$ C - \u0110\u00fang v\u00ec: <br\/> +) Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1 c\u00e2u $A$, ta c\u00f3: <br\/> $AM + CP > \\dfrac{3b}{2}$ (3) <br\/> $AM + BN > \\dfrac{3c}{2}$ (4) <br\/> C\u1ed9ng t\u1eebng v\u1ebf c\u1ee7a (1), (3), (4) ta \u0111\u01b0\u1ee3c: <br\/> $2(AM + BN + CP) > \\dfrac{3a + 3b + 3c}{2}$ <br\/> Hay $AM + BN + AC > \\dfrac{3}{4}(a + b + c)$ (*) <br\/> +) Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1 c\u00e2u $B$ ta c\u0169ng c\u00f3: <br\/> $BN < \\dfrac{a + c}{2}$ (5) <br\/> $CP < \\dfrac{a + b}{2}$ (6) <br\/> C\u1ed9ng theo v\u1ebf c\u1ee7a (2), (5), (6) ta \u0111\u01b0\u1ee3c: <br\/> $AM + BN + CP < \\dfrac{2a + 2b + 2c}{2}$ <br\/> $\\Leftrightarrow$ $AM + BN + CP < a + b + c$ (**) <br\/> T\u1eeb (*) v\u00e0 (**) $\\Rightarrow$ $\\dfrac{3}{4}(a + b + c) < AM + BN + CP < a + b + c$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: D <\/span> <br\/> <span class='basic_green'> <i> Nh\u1eadn x\u00e9t: Ta v\u1eeba ch\u1ee9ng minh \u0111\u01b0\u1ee3c trong m\u1ed9t tam gi\u00e1c <br\/> +) \u0110\u1ed9 d\u00e0i \u0111\u01b0\u1eddng trung tuy\u1ebfn \u1ee9ng v\u1edbi m\u1ed9t c\u1ea1nh l\u1edbn h\u01a1n hi\u1ec7u \u0111\u1ed9 d\u00e0i hai c\u1ea1nh c\u00f2n l\u1ea1i nh\u01b0ng nh\u1ecf h\u01a1n t\u1ed5ng \u0111\u1ed9 d\u00e0i hai c\u1ea1nh c\u00f2n l\u1ea1i c\u1ee7a tam gi\u00e1c <br\/> +) T\u1ed5ng \u0111\u1ed9 d\u00e0i ba \u0111\u01b0\u1eddng trung tuy\u1ebfn l\u1edbn h\u01a1n $\\dfrac{3}{4}$ chu vi nh\u01b0ng nh\u1ecf h\u01a1n chu vi tam gi\u00e1c <\/i><\/span> ","column":1}]}],"id_ques":1915},{"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 $ABC$, ba \u0111\u01b0\u1eddng trung tuy\u1ebfn $AD; BE; CF$ c\u1eaft nhau t\u1ea1i $O$ <br\/> Trong c\u00e1c kh\u1eb3ng \u0111\u1ecbnh sau kh\u1eb3ng \u0111\u1ecbnh n\u00e0o \u0111\u00fang. ","select":["<span class='basic_left'> A. $S_{\\triangle{OAE}} = S_{\\triangle{OEC}} = S_{\\triangle{OCD}} $ <\/span> "," <span class='basic_left'> B. $S_{\\triangle{OCD}} = S_{\\triangle{OBD}} = S_{\\triangle{OFA}} $ <\/span>"," <span class='basic_left'> C. $S_{\\triangle{OBF}} = S_{\\triangle{OFA}} = S_{\\triangle{OAE}}$ <\/span>"," <span class='basic_left'> D. C\u1ea3 3 \u0111\u00e1p \u00e1n tr\u00ean \u0111\u1ec1u \u0111\u00fang <\/span>"],"hint":"So s\u00e1nh di\u1ec7n t\u00edch t\u1eebng tam gi\u00e1c nh\u1ecf v\u1edbi di\u1ec7n t\u00edch tam gi\u00e1c $ABC$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> So s\u00e1nh $S_{\\triangle{OAE}}$ v\u1edbi $S_{\\triangle{ABE}}$ b\u1eb1ng c\u00e1ch so s\u00e1nh chi\u1ec1u cao v\u00e0 c\u1ea1nh \u0111\u00e1y <br\/> <b> B\u01b0\u1edbc 2: <\/b> So s\u00e1nh $S_{\\triangle{ABE}}$ v\u00e0 $S_{\\triangle{ABC}}$ b\u1eb1ng c\u00e1ch t\u01b0\u01a1ng t\u1ef1 <br\/> <b> B\u01b0\u1edbc 3: <\/b> So s\u00e1nh $S_{\\triangle{OAE}}$ v\u00e0 $S_{\\triangle{ABC}}$ <br\/> <b> B\u01b0\u1edbc 4: <\/b> T\u01b0\u01a1ng t\u1ef1 ta so s\u00e1nh di\u1ec7n t\u00edch c\u00e1c tam gi\u00e1c $OEC; OCD; OBD; OBF; OFA$ v\u1edbi di\u1ec7n t\u00edch tam gi\u00e1c $ABC$ v\u00e0 k\u1ebft lu\u1eadn <br\/><br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai20/lv3/img\/H7C3B20_K2.png' \/><\/center> <br\/> $\\blacktriangleright$ V\u00ec $O$ l\u00e0 tr\u1ecdng t\u00e2m c\u1ee7a tam gi\u00e1c $ABC$ n\u00ean: $OE = \\dfrac{1}{3}BE$ (t\u00ednh ch\u1ea5t tr\u1ecdng t\u00e2m) <br\/> X\u00e9t $\\triangle{OAE}$ v\u00e0 $\\triangle{ABE}$ c\u00f3 chung chi\u1ec1u cao h\u1ea1 t\u1eeb $A$ xu\u1ed1ng $BE$ <br\/> C\u1ea1nh \u0111\u00e1y $OE$ b\u1eb1ng $\\dfrac{1}{3}$ c\u1ea1nh \u0111\u00e1y $BE$ <br\/> $\\Rightarrow$ $S_{\\triangle{OAE}} = \\dfrac{1}{3}S_{\\triangle{ABE}}$ (*) <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{ABE}$ v\u00e0 $\\triangle{ABC}$ c\u00f3 chung chi\u1ec1u cao h\u1ea1 t\u1eeb $B$ xu\u1ed1ng $AC$ <br\/> C\u1ea1nh \u0111\u00e1y $AE$ b\u1eb1ng $\\dfrac{1}{2}$ c\u1ea1nh \u0111\u00e1y $AC$ (v\u00ec $BE$ l\u00e0 trung tuy\u1ebfn \u1ee9ng v\u1edbi c\u1ea1nh $AC$) <br\/> $\\Rightarrow$ $S_{\\triangle{ABE}} = \\dfrac{1}{2}S_{\\triangle{ABC}}$ (**) <br\/> $\\blacktriangleright$ T\u1eeb (*) v\u00e0 (**) $\\Rightarrow$ $S_{OAE} = \\dfrac{1}{3} . \\dfrac{1}{2}S_{\\triangle{ABC}} = \\dfrac{1}{6}S_{\\triangle{ABC}}$ (1) <br\/> $\\blacktriangleright$ L\u1eadp lu\u1eadn t\u01b0\u01a1ng t\u1ef1 ta c\u00f3: <br\/> $S_{\\triangle{OFA}} = \\dfrac{1}{6}S_{\\triangle{ABC}}$ (2) <br\/> $S_{\\triangle{OBF}} = \\dfrac{1}{6}S_{\\triangle{ABC}}$ (3) <br\/> $S_{\\triangle{ODB}} = \\dfrac{1}{6}S_{\\triangle{ABC}}$ (4) <br\/> $S_{\\triangle{OCD}} = \\dfrac{1}{6}S_{\\triangle{ABC}}$ (5) <br\/> $S_{\\triangle{OEC}} = \\dfrac{1}{6}S_{\\triangle{ABC}}$ (6) <br\/> $\\blacktriangleright$ T\u1eeb (1), (2), (3), (4), (5), (6) $\\Rightarrow$ $S_{\\triangle{OAE}} = S_{\\triangle{OEC}} = S_{\\triangle{OCD}} = S_{\\triangle{OBD}} = S_{\\triangle{OBF}} = S_{\\triangle{OFA}} = \\dfrac{1}{6}S_{\\triangle{ABC}}$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: D <\/span><br\/> <span class='basic_green'> <i> Nh\u1eadn x\u00e9t: Ba \u0111\u01b0\u1eddng trung tuy\u1ebfn c\u1ee7a tam gi\u00e1c chia tam gi\u00e1c th\u00e0nh $6$ tam gi\u00e1c nh\u1ecf c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng nhau v\u00e0 b\u1eb1ng $\\dfrac{1}{6}$ di\u1ec7n t\u00edch tam gi\u00e1c ban \u0111\u1ea7u <br\/> +) \u1ede b\u00e0i to\u00e1n tr\u00ean ta \u0111\u00e3 s\u1eed d\u1ee5ng ph\u01b0\u01a1ng ph\u00e1p so s\u00e1nh di\u1ec7n t\u00edch b\u1eb1ng c\u00e1ch so s\u00e1nh chi\u1ec1u cao v\u00e0 c\u1ea1nh \u0111\u00e1y. <\/i><\/span> ","column":1}]}],"id_ques":1916},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"Cho tam gi\u00e1c $ABC$, tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia $BC$ l\u1ea5y \u0111i\u1ec3m $E$, tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia $CB$ l\u1ea5y \u0111i\u1ec3m $F$ sao cho $BE = CF$. \u0110i\u1ec3m $G$ l\u00e0 tr\u1ecdng t\u00e2m c\u1ee7a tam gi\u00e1c $ABC$, n\u1ed1i $AG$ c\u1eaft $BC$ t\u1ea1i $M$. L\u1ea5y $H$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n $AG$. N\u1ed1i $EG$ c\u1eaft $AF$ t\u1ea1i $N$ v\u00e0 l\u1ea5y $I$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n $EG$. Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y \u0111\u00fang? ","select":["A. $\\triangle{ABC}$ v\u00e0 $\\triangle{AEF}$ kh\u00f4ng c\u00f9ng tr\u1ecdng t\u00e2m ","B. $IH = MN; IH \/\/ MN$","C. $IH > MN$ ","D. $H$ l\u00e0 tr\u1ecdng t\u00e2m tam gi\u00e1c $AEF$"],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> H\u01b0\u1edbng gi\u1ea3i: V\u1ebd h\u00ecnh ch\u00ednh x\u00e1c gi\u00fap ta ph\u00e1t hi\u1ec7n $G$ c\u0169ng l\u00e0 tr\u1ecdng t\u00e2m tam gi\u00e1c $AEF$ <br\/> <b> B\u01b0\u1edbc 1: <\/b> Ch\u1ee9ng minh $G$ l\u00e0 tr\u1ecdng t\u00e2m tam gi\u00e1c $AEF$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> X\u00e9t $\\triangle{GHI}$ v\u00e0 $\\triangle{GMN}$ \u0111\u1ec3 so s\u00e1nh $IH$ v\u00e0 $MN$ <br\/><br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai20/lv3/img\/H7C3B20_K4.png' \/><\/center> <br\/> $\\blacktriangleright$ A - Sai v\u00ec: <br\/> $AM$ l\u00e0 trung tuy\u1ebfn c\u1ee7a tam gi\u00e1c $ABC$ n\u00ean $BM = MC$ (1) <br\/> Ta c\u00f3: $BE = CF$ (gi\u1ea3 thi\u1ebft) (2) <br\/> T\u1eeb (1) v\u00e0 (2), ta c\u00f3: $BE + BM = MC + CF$ hay $EM = MF$ <br\/> $\\Rightarrow$ $AM$ c\u0169ng l\u00e0 trung tuy\u1ebfn thu\u1ed9c c\u1ea1nh $EF$ c\u1ee7a tam gi\u00e1c $AEF$ <br\/> M\u00e0 $AG = \\dfrac{2}{3}AM$ ($G$ l\u00e0 tr\u1ecdng t\u00e2m c\u1ee7a tam gi\u00e1c $ABC$) <br\/> $\\Rightarrow$ $G$ c\u0169ng l\u00e0 tr\u1ecdng t\u00e2m c\u1ee7a tam gi\u00e1c $AEF$ hay <b> $\\triangle{ABC}$ v\u00e0 $\\triangle{AEF}$ c\u00f3 c\u00f9ng tr\u1ecdng t\u00e2m $G$ <\/b> <br\/> $\\blacktriangleright$ B - \u0110\u00fang v\u00ec: <br\/> C\u00f3: $HG = \\dfrac{1}{2}AG$ (gt) m\u00e0 $AG = \\dfrac{2}{3}AM$ (t\u00ednh ch\u1ea5t tr\u1ecdng t\u00e2m) <br\/> $\\Rightarrow$ $HG = \\dfrac{1}{2} . \\dfrac{2}{3}AM = \\dfrac{1}{3}AM$ <br\/> M\u1eb7t kh\u00e1c $GM = \\dfrac{1}{3}AM$ (t\u00ednh ch\u1ea5t tr\u1ecdng t\u00e2m) n\u00ean $GM = HG (= \\dfrac{1}{3}AM)$ <br\/> V\u00ec $G$ l\u00e0 tr\u1ecdng t\u00e2m $\\triangle{AEF}$ n\u00ean $EN$ l\u00e0 trung tuy\u1ebfn <br\/> $\\Rightarrow$ $GN = \\dfrac{1}{3}EN$ (t\u00ednh ch\u1ea5t tr\u1ecdng t\u00e2m) (*)<br\/> M\u1eb7t kh\u00e1c $IG = \\dfrac{1}{2}EG$ (gt) v\u00e0 $EG = \\dfrac{2}{3}EN$ n\u00ean $IG = \\dfrac{1}{2} . \\dfrac{2}{3} = \\dfrac{1}{3}EN$ (**) <br\/> T\u1eeb (*) v\u00e0 (**) $\\Rightarrow$ $GN = IG$ <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{GHI}$ v\u00e0 $\\triangle{GMN}$ c\u00f3: <br\/> $\\begin{cases} GM = HG (cmt) \\\\ \\widehat{HGI} = \\widehat{MGN} (\\text{\u0111\u1ed1i} \\hspace{0,2cm} \\text{\u0111\u1ec9nh}) \\\\ GN = IG (cmt) \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{GHI} = \\triangle{GMN}$ (c.g.c) <br\/> $\\Rightarrow$ <b> $IH = MN$ <\/b> (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) <br\/> $\\widehat{H_{1}} = \\widehat{M_{1}}$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) m\u00e0 hai g\u00f3c n\u00e0y \u1edf v\u1ecb tr\u00ed so le trong, b\u1eb1ng nhau n\u00ean <b> $IH \/\/ MN$ <\/b> <br\/> $\\blacktriangleright$ C, D sai v\u00ec $G$ l\u00e0 tr\u1ecdng t\u00e2m $\\triangle{AEF}$ v\u00e0 v\u00ec $B$ \u0111\u00fang <br\/> <span class=\u2019basic_pink\u2019> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: B <\/span>","column":2}]}],"id_ques":1917},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $\\sqrt{2}$","B. $2\\sqrt{2}$","C. $3\\sqrt{2}$"],"ques":"Cho tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i $A$ c\u00f3 $AC = b; AB = c$. Hai trung tuy\u1ebfn $AD$ v\u00e0 $BE$ c\u1eaft nhau \u1edf $G$. T\u00ecm quan h\u1ec7 gi\u1eefa $b$ v\u00e0 $c$ \u0111\u1ec3 $AD \\perp BE$ <br\/> <b> \u0110\u00e1p \u00e1n l\u00e0: <\/b>$b$ = ?$c$ ","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> \u0110\u1eb7t $BC = a$, t\u00ednh $AG$ d\u1ef1a v\u00e0o vi\u1ec7c t\u00ednh $AD$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> T\u00ednh $BE$ d\u1ef1a v\u00e0o tam gi\u00e1c vu\u00f4ng $ABE$ <br\/> <b> B\u01b0\u1edbc 3: <\/b> $AD \\perp BE$ $\\Leftrightarrow$ $AG \\perp GB$ $\\Leftrightarrow$ $AG^2 + GB^2 = AB^2$ <br\/> T\u00ednh $GB$ sau \u0111\u00f3 t\u00ednh $BE$ <br\/> <b> B\u01b0\u1edbc 4: <\/b> T\u1eeb b\u01b0\u1edbc 2 v\u00e0 b\u01b0\u1edbc 3 bi\u1ebfn \u0111\u1ed5i t\u00ecm m\u1ed1i quan h\u1ec7 c\u1ee7a $b$ v\u00e0 $c$ <br\/><br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai20/lv3/img\/H7C3B20_K6.png' \/><\/center> <br\/> $\\blacktriangleright$ \u0110\u1eb7t $BC = a$ <br\/> \u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pitago cho tam gi\u00e1c vu\u00f4ng $ABC$, ta c\u00f3: <br\/> $BC^2 = AB^2 + AC^2$ $\\Leftrightarrow$ $a^2 = b^2 + c^2$ <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{ABC}$ vu\u00f4ng t\u1ea1i $A$, trung tuy\u1ebfn $AD$ \u1ee9ng v\u1edbi c\u1ea1nh huy\u1ec1n $BC$ <br\/> $\\Rightarrow$ $AD = \\dfrac{1}{2}BC = \\dfrac{1}{2}a$ <br\/> V\u00ec $G$ l\u00e0 tr\u1ecdng t\u00e2m n\u00ean ta c\u00f3: $AG = \\dfrac{2}{3}AD = \\dfrac{2}{3}.\\dfrac{1}{2}a = \\dfrac{1}{3}a$ <br\/> V\u00e0 $BG = \\dfrac{2}{3}BE$ <br\/> $\\blacktriangleright$ \u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pitago cho tam gi\u00e1c vu\u00f4ng $ABE$, ta c\u00f3: <br\/> $\\begin{align} BE^2 &= AE^2 + AB^2 \\\\ &= \\left( \\dfrac{AC}{2} \\right) + AB^2 \\\\ &= \\left( \\dfrac{b}{2} \\right)^2 + c^2 \\\\ &= \\dfrac{b^2}{4} + c^2 (1) \\end{align}$ <br\/> $\\blacktriangleright$ \u0110\u1ec3 $AD \\perp BE$ hay $AG \\perp GB$ th\u00ec $AG^2 + GB^2 = AB^2$ <br\/> $\\begin{align} GB^2 &= AB^2 - AG^2 \\\\ &= c^2 - \\left( \\dfrac{a}{3} \\right)^2 \\\\ &= c^2 - \\dfrac{a^2}{9} \\end{align}$ <br\/> M\u1eb7t kh\u00e1c $GB = \\dfrac{2}{3}EB$ n\u00ean $\\left(\\dfrac{2}{3}EB \\right)^2 = c^2 - \\dfrac{a^2}{9}$ <br\/> $\\Leftrightarrow$ $\\dfrac{4}{9}EB^2 = c^2 - \\dfrac{a^2}{9}$ <br\/> $\\Leftrightarrow$ $EB^2 = \\dfrac{9c^2}{4} - \\dfrac{a^2}{4}$ (2) <br\/> T\u1eeb (1) v\u00e0 (2) ta c\u00f3: $\\dfrac{b^2}{4} + c^2 = \\dfrac{9c^2}{4} - \\dfrac{a^2}{4}$ <br\/> $\\Leftrightarrow$ $b^2 + 4c^2 = 9c^2 - a^2$ <br\/> $\\Leftrightarrow$ $b^2 = 5c^2 - a^2$ <br\/> $\\Leftrightarrow$ $b^2 = 5c^2 - (b^2 + c^2)$ <br\/> $\\Leftrightarrow$ $b^2 = 4c^2 - b^2$ <br\/> $\\Leftrightarrow$ $b^2 = 2c^2$ <br\/> $\\Leftrightarrow$ $b^2 = (\\sqrt{2}c)^2$ <br\/> $\\Leftrightarrow$ $b = \\sqrt{2}c$ "}]}],"id_ques":1918},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["14,4"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"Cho tam gi\u00e1c $ABC$ c\u00f3 c\u00e1c \u0111\u01b0\u1eddng trung tuy\u1ebfn $AD = 12cm$; trung tuy\u1ebfn $BE = 9cm$; trung tuy\u1ebfn $CF = 15cm$. T\u00ednh \u0111\u1ed9 d\u00e0i c\u1ea1nh $BC$ (k\u1ebft qu\u1ea3 l\u1ea5y \u0111\u1ebfn ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n th\u1ee9 nh\u1ea5t) <br\/> <b> \u0110\u00e1p \u00e1n l\u00e0: <\/b>$BC \\approx$ _input_ $cm$ ","hint":"H\u01b0\u1edbng gi\u1ea3i: Tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia $DA$ l\u1ea5y \u0111i\u1ec3m $K$ sao cho $DG = DK$ <br\/> V\u1ebd h\u00ecnh ch\u00ednh x\u00e1c gi\u00fap nh\u1eadn th\u1ea5y $\\widehat{BGK}$ vu\u00f4ng","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> Tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia $DA$ l\u1ea5y \u0111i\u1ec3m $K$ sao cho $DG = DK$ <br\/> T\u00ednh $BG; CG; DG; DK$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> Ch\u1ee9ng minh $BK = CG$ <br\/> <b> B\u01b0\u1edbc 3: <\/b> Ch\u1ee9ng minh $\\triangle{GBK}$ vu\u00f4ng t\u1ea1i $G$ b\u1eb1ng \u0111\u1ecbnh l\u00fd \u0111\u1ea3o c\u1ee7a \u0111\u1ecbnh l\u00fd Pitago <br\/> <b> B\u01b0\u1edbc 4: <\/b> T\u00ednh $BD$ r\u1ed3i t\u00ednh $BC$ <br\/><br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai20/lv3/img\/H7C3B20_K7.png' \/><\/center> <br\/> $\\blacktriangleright$ Tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia $DA$ l\u1ea5y \u0111i\u1ec3m $K$ sao cho $DG = DK$ <br\/> G\u1ecdi $G$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a c\u00e1c \u0111\u01b0\u1eddng trung tuy\u1ebfn, do \u0111\u00f3 $G$ l\u00e0 tr\u1ecdng t\u00e2m c\u1ee7a tam gi\u00e1c $ABC$<br\/> V\u00ec $G$ l\u00e0 tr\u1ecdng t\u00e2m c\u1ee7a tam gi\u00e1c $ABC$ n\u00ean theo t\u00ednh ch\u1ea5t tr\u1ecdng t\u00e2m, ta c\u00f3: <br\/> $CG = \\dfrac{2}{3}CF = \\dfrac{2}{3}.15 = 10(cm)$ <br\/> $BG = \\dfrac{2}{3}BE = \\dfrac{2}{3}.9 = 6(cm)$ <br\/> $GA = \\dfrac{2}{3}AD$ $\\Rightarrow$ $GD = \\dfrac{1}{3}AD = \\dfrac{1}{3}.12 = 4(cm)$ <br\/> V\u00ec $GD = DK$ (theo c\u00e1ch l\u1ea5y \u0111i\u1ec3m $K$) $\\Rightarrow$ $GK = 2.GD = 2 . 4 = 8(cm)$ <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{GDC}$ v\u00e0 $\\triangle{KDB}$ c\u00f3: <br\/> $\\begin{cases} BD = DC (gt) \\\\ \\widehat{GDC} = \\widehat{KDB} \\\\ GD = DK \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{GDC} = \\triangle{KDB}$ (c.g.c) <br\/> $\\Rightarrow$ $BK = GC$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{GBK}$ c\u00f3 $3$ c\u1ea1nh: <br\/> $BG = 6cm; GK = 8cm; BK = 10cm$ <br\/> Ta c\u00f3: $10^2 = 6^2 + 8^2$ <br\/> $\\Rightarrow$ $BK^2 = BG^2 + GK^2$ <br\/> $\\Rightarrow$ $\\triangle{GBK}$ vu\u00f4ng t\u1ea1i $G$ (\u0111\u1ecbnh l\u00fd Pitago \u0111\u1ea3o) <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{BGD}$ vu\u00f4ng t\u1ea1i $G$ c\u00f3: <br\/> $\\begin{align} BD^2 &= BG^2 + GD^2 \\\\ &= 6^2 + 4^2 \\\\ &= 52 \\end{align}$ <br\/> $\\Rightarrow$ $BD = \\sqrt{52}$ <br\/> $\\Rightarrow$ $BC = 2BD = 2\\sqrt{52} \\approx 14,4 (cm)$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n l\u00e0: $14,4$ <\/span> "}]}],"id_ques":1919},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00fang hay sai","title_trans":"","temp":"true_false","correct":[["t","t","t","f"]],"list":[{"point":10,"image":"","col_name":["Kh\u1eb3ng \u0111\u1ecbnh","\u0110\u00fang","Sai"],"arr_ques":[" N\u1ebfu $\\triangle{ABC}$ \u0111\u1ec1u v\u00e0 c\u00f3 $G$ l\u00e0 tr\u1ecdng t\u00e2m th\u00ec $GA = GB = GC$ "," $G$ l\u00e0 tr\u1ecdng t\u00e2m tam gi\u00e1c $ABC$. N\u1ebfu c\u00f3 $GA = GB = GC$ th\u00ec $\\triangle{ABC}$ \u0111\u1ec1u "," Tam gi\u00e1c $ABC$ c\u00f3 hai \u0111\u01b0\u1eddng trung tuy\u1ebfn $BM$ v\u00e0 $CN$. N\u1ebfu $AB < AC$ th\u00ec $BM < CN$ ","Tam gi\u00e1c $ABC$ c\u00f3 hai \u0111\u01b0\u1eddng trung tuy\u1ebfn $BM$ v\u00e0 $CN$. N\u1ebfu $BM < CN$ th\u00ec $AB > AC$"],"hint":"","explain":["<span class='basic_left'> \u0110\u00daNG v\u00ec: <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai20/lv3/img\/H7C3B20_K10.png' \/><\/center> <br\/> X\u00e9t $\\triangle{ABC}$ v\u1edbi ba \u0111\u01b0\u1eddng trung tuy\u1ebfn $AM; BN$ v\u00e0 $CP$ c\u1eaft nhau t\u1ea1i $G$ <br\/> Theo t\u00ednh ch\u1ea5t tr\u1ecdng t\u00e2m, ta c\u00f3: <br\/> $GA = \\dfrac{2}{3}AM; GB = \\dfrac{2}{3}BN; GC = \\dfrac{2}{3}CP$ (1) <br\/> $\\blacktriangleright$ Theo gi\u1ea3 thi\u1ebft $\\triangle{ABC}$ \u0111\u1ec1u $\\Rightarrow$ $AB = BC = CA$ <br\/> V\u00e0 $\\widehat{A} = \\widehat{B} = \\widehat{C} = 60^{o}$ <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{ABN}$ v\u00e0 $\\triangle{ACP}$ c\u00f3: <br\/> $\\begin{cases} AB = AC (gt) \\\\ \\widehat{A} \\hspace{0,2cm} \\text{chung} \\\\ AN = AP (\\text{v\u00ec} AB = AC) \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{ABN} = \\triangle{ACP}$ (c.g.c) <br\/> $\\Rightarrow$ $BN = CP$ (2) <br\/> $\\blacktriangleright$ Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1 ta c\u00f3: $AM = BN$ (3) <br\/> T\u1eeb (1); (2) v\u00e0 (3) $\\Rightarrow$ $GA = GB = GC$ <\/span> "," <span class='basic_left'> <br\/> \u0110\u00daNG v\u00ec: <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai20/lv3/img\/H7C3B20_K10.png' \/><\/center> <br\/> Theo gi\u1ea3 thi\u1ebft $GA = GB = GC$ <br\/> M\u1eb7t kh\u00e1c v\u00ec $G$ l\u00e0 tr\u1ecdng t\u00e2m tam gi\u00e1c $ABC$ ta c\u00f3: <br\/> $GA = \\dfrac{2}{3}AM; GB = \\dfrac{2}{3}BN; GC = \\dfrac{2}{3}CP$ <br\/> $\\Rightarrow$ $GM = GN = GP$ <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{GBP}$ v\u00e0 $\\triangle{GCN}$ c\u00f3: <br\/> $\\begin{cases} GB = GC (gt) \\\\ \\widehat{NGC} = \\widehat{PGB} (\\text{\u0111\u1ed1i} \\hspace{0,2cm} \\text{\u0111\u1ec9nh}) \\\\ GP = GN (cmt) \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{GBP} = \\triangle{GCN}$ (c.g.c) <br\/> $\\Rightarrow$ $BP = CN$ (1) <br\/> M\u1eb7t kh\u00e1c $AP = BP; AN = NC$ (V\u00ec $BN; CP$ l\u00e0 trung tuy\u1ebfn) (2) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow$ $AB = AC$ (3) <br\/> Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1 ta c\u00f3: $AC = BC$ (4) <br\/> T\u1eeb (3) v\u00e0 (4) $\\Rightarrow$ $AB = AC = BC$ hay $\\triangle{ABC}$ \u0111\u1ec1u <\/span> ","<span class='basic_left'> <br\/> \u0110\u00daNG v\u00ec: <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai20/lv3/img\/H7C3B20_K11.png' \/><\/center> <br\/> $\\blacktriangleright$ G\u1ecdi $G$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $BM$ v\u00e0 $CN$ <br\/> Theo t\u00ednh ch\u1ea5t tr\u1ecdng t\u00e2m, ta c\u00f3: <br\/> $GB = \\dfrac{2}{3}BM; GC = \\dfrac{2}{3}CN$ <br\/> Tia $AG$ c\u1eaft $BC$ \u1edf $I$, suy ra $IB = IC$ (v\u00ec $AI$ l\u00e0 trung tuy\u1ebfn) <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{AIB}$ v\u00e0 $\\triangle{AIC}$ c\u00f3: <br\/> $\\begin{cases} AI \\hspace{0,2cm} \\text{chung} \\\\ IB = IC (cmt) \\\\ AB < AC (gt) \\end{cases}$ <br\/> $\\Rightarrow$ $\\widehat{AIB} < \\widehat{AIC}$ (\u0111\u1ecbnh l\u00fd b\u1ed5 sung - quan h\u1ec7 gi\u1eefa g\u00f3c v\u00e0 c\u1ea1nh \u0111\u1ed1i di\u1ec7n) <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{GIB}$ v\u00e0 $\\triangle{GIC}$ c\u00f3: <br\/> $\\begin{cases} GI \\hspace{0,2cm} \\text{chung} \\\\ \\widehat{GIB} < \\widehat{GIC} (cmt) \\\\ IB = IC (cmt) \\end{cases}$ <br\/> $\\Rightarrow$ $GB < GC$ (\u0111\u1ecbnh l\u00fd b\u1ed5 sung - quan h\u1ec7 gi\u1eefa g\u00f3c v\u00e0 c\u1ea1nh \u0111\u1ed1i di\u1ec7n) <br\/> Do \u0111\u00f3: $BM < CN$ <\/span> ","<span class='basic_left'> <br\/> SAI v\u00ec: <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai20/lv3/img\/H7C3B20_K11.png' \/><\/center> <br\/> $\\blacktriangleright$ G\u1ecdi $G$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $BM$ v\u00e0 $CN$ <br\/> $AG$ c\u1eaft $BC$ t\u1ea1i $I$ $\\Rightarrow$ $IB = IC$ (do $AI$ l\u00e0 trung tuy\u1ebfn) <br\/> Theo t\u00ednh ch\u1ea5t tr\u1ecdng t\u00e2m, ta c\u00f3: <br\/> $GB = \\dfrac{2}{3}BM; GC = \\dfrac{2}{3}CN$ <br\/> Theo gi\u1ea3 thi\u1ebft $BM < CN$ $\\Rightarrow$ $GB < GC$ <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{GIB}$ v\u00e0 $\\triangle{GIC}$ c\u00f3: <br\/> $\\begin{cases} AI \\hspace{0,2cm} \\text{chung} \\\\ IB = IC (cmt) \\\\ Gb < GC (gt) \\end{cases}$ <br\/> $\\Rightarrow$ $\\widehat{GIB} < \\widehat{GIC}$ (\u0111\u1ecbnh l\u00fd b\u1ed5 sung - quan h\u1ec7 gi\u1eefa g\u00f3c v\u00e0 c\u1ea1nh \u0111\u1ed1i di\u1ec7n) <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{AIB}$ v\u00e0 $\\triangle{AIC}$ c\u00f3: <br\/> $\\begin{cases} AI \\hspace{0,2cm} \\text{chung} \\\\ IB = IC (cmt) \\\\ \\widehat{AIB} < \\widehat{AIC} (cmt) \\end{cases}$ <br\/> $\\Rightarrow$ $AB < AC$ (\u0111\u1ecbnh l\u00fd b\u1ed5 sung - quan h\u1ec7 gi\u1eefa g\u00f3c v\u00e0 c\u1ea1nh \u0111\u1ed1i di\u1ec7n) <\/span> "]}]}],"id_ques":1920}],"lesson":{"save":0,"level":3}}