{"segment":[{"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='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai21/lv3/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>"}]}],"id_ques":1941},{"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='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai21/lv3/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>"}]}],"id_ques":1942},{"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='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai21/lv3/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}]}],"id_ques":1943},{"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":[[["3"]]],"list":[{"point":10,"width":50,"content":"","type_input":"","ques":"Cho tam gi\u00e1c $ABC$ c\u00f3 $\\widehat{A} = 90^{o}; AB = 8cm; AC = 15cm$. G\u1ecdi $O$ l\u00e0 giao \u0111i\u1ec3m c\u00e1c tia ph\u00e2n gi\u00e1c c\u1ee7a $\\triangle{ABC}$. T\u00ednh kho\u1ea3ng c\u00e1ch t\u1eeb $O$ \u0111\u1ebfn c\u00e1c c\u1ea1nh c\u1ee7a tam gi\u00e1c $ABC$. <br\/> <b> \u0110\u00e1p \u00e1n l\u00e0: <\/b> _input_$cm$ ","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 $BC$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> Ch\u1ee9ng minh $OD = OE = OI = AD = AE$<br\/> <b> B\u01b0\u1edbc 3: <\/b> T\u00ednh $AD$ b\u1eb1ng c\u00e1ch quy v\u1ec1 c\u00e1c c\u1ea1nh $AB, AC, BC$ \u0111\u00e3 bi\u1ebft <br\/> C\u1ee5 th\u1ec3: $AD = AB - BD$; $AE = AC - EC$ <br\/> Ch\u1ee9ng minh: $BD = BI; CI = CE$ <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai21/lv3/img\/H7C3B21_K02.png' \/><\/center> $\\blacktriangleright$ \u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pitago cho $\\triangle{ABC}$ vu\u00f4ng t\u1ea1i $A$, ta c\u00f3: <br\/> $\\begin{align} BC^2 &= AB^2 + AC^2 \\\\ & = 8^2 + 15^2 \\\\ &= 64 + 225 \\\\ &= 289 \\end{align}$ <br\/> $\\Rightarrow$ $BC = 17(cm)$ <br\/> $\\blacktriangleright$ T\u1eeb $O$ k\u1ebb $OD \\perp AB; OI \\perp BC; OE \\perp AC$ <br\/> V\u00ec $O$ l\u00e0 giao \u0111i\u1ec3m c\u00e1c tia ph\u00e2n gi\u00e1c c\u1ee7a $\\triangle{ABC}$ <br\/> $\\Rightarrow$ $OI = OD = OE$ (t\u00ednh ch\u1ea5t ba \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c trong tam gi\u00e1c) (1) <br\/> V\u00ec $AO$ l\u00e0 tia ph\u00e2n gi\u00e1c $\\widehat{A}$ m\u00e0 $\\widehat{A} = 90^{o}$ <br\/> $\\Rightarrow$ $\\widehat{DAO} = \\widehat{EAO} = \\dfrac{\\widehat{A}}{2} = \\dfrac{90^{o}}{2} = 45^{o}$ <br\/>C\u00e1c tam gi\u00e1c vu\u00f4ng $ADO$ v\u00e0 $AEO$ c\u00f3 $\\widehat{DAO} = 45^{o}$ v\u00e0 $\\widehat{EAO} = 45^{o}$ <br\/> $\\Rightarrow$ $\\triangle{DAO}$ v\u00e0 $\\triangle{EAO}$ vu\u00f4ng c\u00e2n t\u1ea1i $D$ v\u00e0 $E$ <br\/> $\\Rightarrow$ $AD = DO$ v\u00e0 $AE = EO$ (\u0111\u1ecbnh ngh\u0129a tam gi\u00e1c c\u00e2n) (2) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow$ $OD = OE = OI = AD = AE$ <br\/> $\\blacktriangleright$ X\u00e9t hai tam gi\u00e1c vu\u00f4ng $ODB$ v\u00e0 $OIB$ c\u00f3: <br\/> $\\begin{cases} OD = OI (cmt) \\\\ BO \\hspace{0,2cm} \\text{chung} \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{OBD} = \\triangle{OID}$ (c\u1ea1nh huy\u1ec1n - c\u1ea1nh g\u00f3c vu\u00f4ng) <br\/> $\\Rightarrow$ $BD = BI$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) <br\/> Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1 ta c\u00f3: $CI = CE$ <br\/> $\\blacktriangleright$ V\u00ec $AD = AB - BD = AB - BI$ <br\/> $AE = AC - EC = AC - IC$ <br\/> V\u1eady $AD + AE = (AB - BI) + (AC - IC) = AB + AC - (BI + IC) = AB + AC - BC$ <br\/> V\u00ec $AD = AE$ $\\Rightarrow$ $2AD = AB + AC - BC$ <br\/> Do \u0111\u00f3: $AD = \\dfrac{AB + AC - BC}{2} = \\dfrac{8 + 15 - 17}{2} = 3(cm)$ <br\/> $\\Rightarrow$ $OD = OE = OI = AD = 3cm$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p c\u1ea7n \u0111i\u1ec1n l\u00e0: $3$ <\/span>"}]}],"id_ques":1944},{"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 g\u00f3c $xOy$, tr\u00ean $Ox$ l\u1ea5y \u0111i\u1ec3m $A$ v\u00e0 $B$, tr\u00ean $Oy$ l\u1ea5y hai \u0111i\u1ec3m $C$ v\u00e0 $D$ sao cho $OA = OC; OB = OD$. T\u1eeb $A$ v\u00e0 $B$ k\u1ebb $An \/\/ Oy, Bn' \/\/ Oy$. T\u1eeb $C$ v\u00e0 $D$ k\u1ebb $Cm \/\/ Ox, Dm' \/\/ Ox$. Bi\u1ebft $An$ v\u00e0 $Cm$ c\u1eaft nhau t\u1ea1i $E$, $Bn'$ v\u00e0 $Dm'$ c\u1eaft nhau t\u1ea1i $F$. Khi \u0111\u00f3 $E, O, F$ th\u1eb3ng h\u00e0ng. <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> Ch\u1ee9ng minh $OE$ l\u00e0 tia ph\u00e2n gi\u00e1c $\\widehat{xOy}$ b\u1eb1ng c\u00e1ch ch\u1ee9ng minh $\\widehat{O_{1}} = \\widehat{O_{2}}$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> Ch\u1ee9ng minh $OF$ l\u00e0 tia ph\u00e2n gi\u00e1c $\\widehat{xOy}$ b\u1eb1ng c\u00e1ch t\u01b0\u01a1ng t\u1ef1 <br\/><br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span> <br\/> <img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai21/lv3/img\/H7C3B21_K03A.png' \/> <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai21/lv3/img\/H7C3B21_K03.png' \/><\/center> <br\/> <br\/> $\\blacktriangleright$ N\u1ed1i $OE$, x\u00e9t $\\triangle{OAE}$ v\u00e0 $\\triangle{OCE}$ c\u00f3: <br\/> $\\begin{cases} OE \\hspace{0,2cm} \\text{chung} \\\\ \\widehat{O_{1}} = \\widehat{E_{2}} (\\text{so} \\hspace{0,2cm} \\text{le} \\hspace{0,2cm} \\text{trong}) (1) \\\\ \\widehat{O_{2}} = \\widehat{E_{1}} (\\text{so} \\hspace{0,2cm} \\text{le} \\hspace{0,2cm} \\text{trong}) (2) \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{OAE} = \\triangle{ECO}$ (g.c.g) $\\Rightarrow$ $AE = CO$ <br\/> M\u1eb7t kh\u00e1c ta c\u00f3: $OA = OC$ (gt) $\\Rightarrow$ $OA = AE$ <br\/> $\\Rightarrow$ $\\triangle{OAE}$ c\u00e2n t\u1ea1i $A$, suy ra $\\widehat{O_{1}} = \\widehat{E_{1}}$ (t\u00ednh ch\u1ea5t tam gi\u00e1c c\u00e2n) (3) <br\/> T\u1eeb (2), (3) $\\Rightarrow$ $\\widehat{O_{1}} = \\widehat{O_{2}}$ <br\/> $\\Rightarrow$ $OE$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{xOy}$ (*) <br\/> $\\blacktriangleright$ N\u1ed1i $OF$, t\u01b0\u01a1ng t\u1ef1 x\u00e9t $\\triangle{BOF}$ v\u00e0 $\\triangle{DOF}$ c\u00f3: <br\/> $\\begin{cases} \\widehat{BFO} = \\widehat{FOD} (\\text{so} \\hspace{0,2cm} \\text{le} \\hspace{0,2cm} \\text{trong}) (4) \\\\ OF \\hspace{0,2cm} \\text{chung} \\\\ \\widehat{DFO} = \\widehat{BOF} (\\text{so} \\hspace{0,2cm} \\text{le} \\hspace{0,2cm} \\text{trong}) \\end{cases}$ <br\/> V\u1eady $\\triangle{OBF} = \\triangle{FDO}$ (g.c.g) <br\/> $\\Rightarrow$ $OD = BF$ m\u00e0 $OB = OD$ $\\Rightarrow$ $OB = BF$ <br\/> V\u1eady $\\triangle{BOF}$ c\u00e2n t\u1ea1i $B$ $\\Rightarrow$ $\\widehat{BFO} = \\widehat{BOF}$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) (5) <br\/> T\u1eeb (4) v\u00e0 (5) $\\Rightarrow$ $\\widehat{BOF} = \\widehat{FOD}$ <br\/> V\u1eady $OF$ l\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $xOy$ (**) <br\/> T\u1eeb (*) v\u00e0 (**) suy ra $OE; OF$ c\u00f9ng l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{xOy}$ <br\/> V\u1eady $OE$ tr\u00f9ng $OF$ hay $O; E; F$ th\u1eb3ng h\u00e0ng <br\/> V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 <span class='basic_pink'> \u0110\u00daNG <\/span> <br\/> <span class='basic_green'> <i> Sai l\u1ea7m d\u1ec5 m\u1eafc ph\u1ea3i \u1edf ch\u1ed7: <br\/> +) Ch\u1ee9ng minh xong $\\triangle{OAE} = \\triangle{ECO}$ l\u00e0 k\u1ebft lu\u1eadn ngay $\\widehat{O_{1}} = \\widehat{O_{2}}$ \u0111\u1ec3 suy ra $OE$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{xOy}$ l\u00e0 sai, v\u00ec hai tam gi\u00e1c tr\u00ean b\u1eb1ng nhau th\u00ec$\\widehat{O_{2}}$ t\u01b0\u01a1ng \u1ee9ng v\u1edbi $\\widehat{E_{1}}$ (ch\u1ee9 kh\u00f4ng t\u01b0\u01a1ng \u1ee9ng v\u1edbi $\\widehat{O_{1}}$) <br\/> +) Ta d\u1ec5 d\u00e0ng th\u1ea5y $OA \\neq OC$ th\u00ec $\\triangle{OAE}$ v\u00e0 $\\triangle{ECO}$ v\u1eabn b\u1eb1ng nhau theo tr\u01b0\u1eddng h\u1ee3p g.c.g <br\/> Nh\u01b0ng ta d\u1ec5 d\u00e0ng th\u1ea5y $\\widehat{E_{1}} = \\widehat{O_{2}}$ v\u00e0 $\\widehat{E_{2}} = \\widehat{O_{1}}$ v\u00e0 $\\widehat{O_{1}} \\neq \\widehat{O_{2}}$ n\u00ean $OE$ kh\u00f4ng ph\u1ea3i tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $xOy$ <\/i> <\/span> ","column":2}]}],"id_ques":1945},{"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 $ABC$, tia ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{B}$ v\u00e0 $\\widehat{C}$ c\u1eaft nhau t\u1ea1i $O$, bi\u1ebft $\\widehat{A} = 70^{o}$. Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y \u0111\u00fang? ","select":["A. $AO$ l\u00e0 tia ph\u00e2n gi\u00e1c $\\widehat{A}$ ","B. $\\widehat{BOC} = 125^{o}$ ","C. $\\widehat{BOC} = 120^{o}$ ","D. A v\u00e0 B \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> Ch\u1ee9ng minh $OK = OI$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> D\u1ef1a v\u00e0o \u0111\u1ecbnh l\u00fd t\u1ed5ng ba g\u00f3c t\u00ednh $\\widehat{BOC}$ <br\/><br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span> <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai21/lv3/img\/H7C3B21_K04.png' \/><\/center> <br\/> <br\/> $\\blacktriangleright$ H\u1ea1 $OK \\perp AB; OI \\perp AC; OH \\perp BC$ <br\/> V\u00ec \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c g\u00f3c $B$ v\u00e0 g\u00f3c $C$ c\u1eaft nhau t\u1ea1i $O$ n\u00ean $O$ l\u00e0 giao \u0111i\u1ec3m ba \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c trong $\\triangle{ABC}$ <br\/> $\\Rightarrow$ $AO$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{A}$ <br\/>$\\blacktriangleright$ $\\triangle{ABC}$ c\u00f3 $BO$ v\u00e0 $CO$ l\u00e0 hai tia ph\u00e2n gi\u00e1c n\u00ean: <br\/> $\\widehat{B_{1}} = \\widehat{B_{2}} = \\dfrac{1}{2}\\widehat{B}$; $\\widehat{C_{1}} = \\widehat{C_{2}} = \\dfrac{1}{2}\\widehat{C}$ <br\/> $\\Rightarrow$ $\\widehat{B_{2}} + \\widehat{C_{2}} = \\dfrac{1}{2}(\\widehat{B} + \\widehat{C})$ (1) <br\/> $\\blacktriangleright$ $\\triangle{BOC}$ c\u00f3 $\\widehat{BOC} = 180^{o} - (\\widehat{B_{1}} + \\widehat{C_{1}})$ (\u0111\u1ecbnh l\u00fd t\u1ed5ng ba g\u00f3c trong tam gi\u00e1c) (2) <br\/> M\u00e0 $\\widehat{A} + \\widehat{B} + \\widehat{C} = 180^{o}$ (\u0111\u1ecbnh l\u00fd t\u1ed5ng ba g\u00f3c trong $\\triangle{ABC}$) <br\/> $\\Rightarrow$ $\\widehat{B} + \\widehat{C} = 180^{o} - \\widehat{A} = 180^{o} - 70^{o} = 110^{o}$ (3) <br\/> Thay (1) v\u00e0 (2), ta c\u00f3: $\\widehat{BOC} = 180^{o} - \\dfrac{1}{2}(\\widehat{B} + \\widehat{C})$ (4) <br\/> Thay (3) v\u00e0 (4), ta c\u00f3: $\\widehat{BOC} = 180^{o} - \\dfrac{1}{2}.110^{o} = 125^{o}$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: D <\/span> ","column":2}]}],"id_ques":1946},{"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":[[["70"]]],"list":[{"point":10,"width":50,"content":"","type_input":"","ques":"Cho tam gi\u00e1c $ABC$, c\u00f3 $\\widehat{A} = 120^{o}$, c\u00e1c \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c $AD, BE, CF$. T\u00ednh chu vi $\\triangle{DEF}$ bi\u1ebft $DE = 21cm; DF = 20cm$ <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai21/lv3/img\/H7C3B21_K06.png' \/><\/center> <br\/> <b> \u0110\u00e1p \u00e1n l\u00e0: <\/b> _input_ $cm$ ","hint":"Ch\u1ee9ng minh $\\triangle{DEF}$ vu\u00f4ng t\u1ea1i $D$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> C\u1ea7n ch\u1ee9ng minh $DE \\perp DF$ b\u1eb1ng c\u00e1ch ch\u1ee9ng minh $DE$ v\u00e0 $DF$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a hai g\u00f3c k\u1ec1 b\u00f9 sau \u0111\u00f3 t\u00ednh $EF$ r\u1ed3i t\u00ednh chu vi $\\triangle{DEF}$ c\u1ee5 th\u1ec3: <br\/> <b> B\u01b0\u1edbc 1: <\/b> Ch\u1ee9ng minh $DE$ l\u00e0 tia ph\u00e2n gi\u00e1c g\u00f3c ngo\u00e0i t\u1ea1i $D$ c\u1ee7a tam gi\u00e1c $ABD$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> Ch\u1ee9ng minh $DF$ l\u00e0 tia ph\u00e2n gi\u00e1c g\u00f3c ngo\u00e0i t\u1ea1i $D$ c\u1ee7a tam gi\u00e1c $ADC$ <br\/> <b> B\u01b0\u1edbc 3: <\/b> T\u00ednh $EF$ r\u1ed3i t\u00ednh chu vi tam gi\u00e1c $DEF$ <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai21/lv3/img\/H7C3B21_K06.png' \/><\/center> <br\/> $\\blacktriangleright$ V\u00ec $AD$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{A}$ <br\/> N\u00ean $\\widehat{A_{1}} = \\widehat{A_{2}} = \\dfrac{\\widehat{A}}{2} = \\dfrac{120^{o}}{2} = 60^{o}$ (1) <br\/> M\u1eb7t kh\u00e1c $\\widehat{BAC}$ v\u00e0 $\\widehat{CAx}$ l\u00e0 hai g\u00f3c k\u1ec1 b\u00f9 n\u00ean $\\widehat{CAx} = 180^{o} - \\widehat{BAC} = 180^{o} - 120^{o} = 60^{o}$ (2) <br\/> $\\widehat{A_{4}} = \\widehat{A_{3}} = 60^{o}$ (\u0111\u1ed1i \u0111\u1ec9nh) (3) <br\/> T\u1eeb (1), (2), (3) $\\Rightarrow$ $\\widehat{A_{1}} = \\widehat{A_{2}} = \\widehat{A_{3}} = \\widehat{A_{4}} = 60^{o}$ <br\/> X\u00e9t $\\triangle{ABD}$ c\u00f3 $\\widehat{DAx}$ l\u00e0 g\u00f3c ngo\u00e0i t\u1ea1i \u0111\u1ec9nh $A$ <br\/> C\u00f3: $\\widehat{A_{2}} = \\widehat{A_{3}} = 60^{o}$ (cmt) <br\/> $\\Rightarrow$ $AC$ l\u00e0 tia ph\u00e2n gi\u00e1c $\\widehat{DAx}$ <br\/> $BE$ l\u00e0 tia ph\u00e2n gi\u00e1c g\u00f3c trong t\u1ea1i \u0111\u1ec9nh $B$ (gt) <br\/> Hai tia ph\u00e2n gi\u00e1c n\u00e0y c\u1eaft nhau t\u1ea1i $E$ suy ra $DE$ l\u00e0 tia ph\u00e2n gi\u00e1c g\u00f3c ngo\u00e0i t\u1ea1i $D$ c\u1ee7a tam gi\u00e1c $ABD$ (*) <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{ADC}$, l\u1eadp lu\u1eadn t\u01b0\u01a1ng t\u1ef1 ta c\u00f3 $DF$ l\u00e0 tia ph\u00e2n gi\u00e1c g\u00f3c ngo\u00e0i t\u1ea1i \u0111\u1ec9nh $D$ (**) <br\/> T\u1eeb (*) v\u00e0 (**) suy ra $DE \\perp DF$ (tia ph\u00e2n gi\u00e1c c\u1ee7a hai g\u00f3c k\u1ec1 b\u00f9) <br\/> $\\blacktriangleright$ \u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pitago cho tam gi\u00e1c vu\u00f4ng $DEF$ ta \u0111\u01b0\u1ee3c: <br\/> $EF^2 = DE^2 + DF^2 = 21^2 + 20^2 = 841$ <br\/> $\\Rightarrow$ $EF = 29(cm)$ <br\/> Chu vi tam gi\u00e1c $DEF$ l\u00e0: $21 + 20 + 29 = 70(cm)$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p c\u1ea7n \u0111i\u1ec1n l\u00e0: $70$ <\/span> <br\/> <span class='basic_green'> <i> Nh\u1eadn x\u00e9t: \u0110\u1ec3 ch\u1ee9ng minh m\u1ed9t tia l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c ta c\u00f3 th\u1ec3: <br\/> +) D\u00f9ng \u0111\u1ecbnh ngh\u0129a: Ch\u1ee9ng minh tia n\u00e0y n\u1eb1m gi\u1eefa hai c\u1ea1nh c\u1ee7a g\u00f3c v\u00e0 t\u1ea1o v\u1edbi hai c\u1ea1nh \u0111\u00f3 hai g\u00f3c b\u1eb1ng nhau <br\/> +) D\u00f9ng t\u00ednh ch\u1ea5t: Ch\u1ee9ng minh m\u1ed9t \u0111i\u1ec3m tr\u00ean tia n\u00e0y c\u00e1ch \u0111\u1ec1u hai c\u1ea1nh c\u1ee7a g\u00f3c <br\/> +) D\u00f9ng t\u00ednh ch\u1ea5t ba \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c (ho\u1eb7c hai tia ph\u00e2n gi\u00e1c ngo\u00e0i v\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c trong kh\u00f4ng k\u1ec1) c\u1ee7a tam gi\u00e1c c\u00f9ng \u0111i qua m\u1ed9t \u0111i\u1ec3m <br\/> \u1ede b\u00e0i to\u00e1n n\u00e0y ta \u0111\u00e3 v\u1eadn d\u1ee5ng c\u00e1ch th\u1ee9 ba n\u00e0y <\/i> <\/span> "}]}],"id_ques":1947},{"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 $MNP$ c\u00f3 $\\widehat{N} = 45^{o}$, \u0111\u01b0\u1eddng cao $MH$, ph\u00e2n gi\u00e1c $NK$. Cho bi\u1ebft $\\widehat{NKM} = 45^{o}$. Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y \u0111\u00fang? ","select":["A. $\\widehat{KHP} = 45^{o}$ ","B. $MN \/\/ KH$ ","C. $\\widehat{M_{1}} = \\widehat{M_{2}}$ ","D. C\u1ea3 $3$ \u0111\u00e1p \u00e1n tr\u00ean \u0111\u1ec1u \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 $\\widehat{M_{1}}$ d\u1ef1a v\u00e0o g\u00f3c ngo\u00e0i c\u1ee7a $\\triangle{MNP}$ v\u00e0 $\\widehat{M_{2}}$ d\u1ef1a v\u00e0o tam gi\u00e1c vu\u00f4ng $MHP$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> T\u00ednh $\\widehat{KHP}$ b\u1eb1ng c\u00e1ch ch\u1ec9 ra $HK$ l\u00e0 tia ph\u00e2n gi\u00e1c $\\widehat{MHP}$ <br\/> <b> B\u01b0\u1edbc 3: <\/b> Ch\u1ee9ng minh $MN \/\/ HK$ <br\/><br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span> <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai21/lv3/img\/H7C3B21_K07.png' \/><\/center> <br\/> <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{KNP}$ c\u00f3 g\u00f3c $\\widehat{MKN}$ l\u00e0 g\u00f3c ngo\u00e0i n\u00ean $ \\widehat{MKN} = \\widehat{N_{2}} + \\widehat{P}$ <br\/> Suy ra $\\widehat{P} = \\widehat{MKN} - \\widehat{N_{2}}$ <br\/> $\\widehat{P} = 45^{o} - \\widehat{N_{2}}$ <br\/> $\\widehat{P} = 45^{o} - \\dfrac{\\widehat{N}}{2}$ <br\/> X\u00e9t $\\triangle{MNP}$ c\u00f3: $\\widehat{M_{1}}$ l\u00e0 g\u00f3c ngo\u00e0i n\u00ean: <br\/> $\\widehat{M_{1}} = \\widehat{N} + \\widehat{P} = \\widehat{N} + (45^{o} - \\dfrac{\\widehat{N}}{2})$ <br\/> $\\Rightarrow$ $\\widehat{M_{1}} = \\dfrac{\\widehat{N}}{2} + 45^{o}$ (1) <br\/> X\u00e9t $\\triangle{MHP}$ vu\u00f4ng t\u1ea1i $H$ c\u00f3: <br\/> $\\widehat{M_{2}} = 90^{o} - \\widehat{P} = 90^{o} - \\left( 45^{o} - \\dfrac{\\widehat{N}}{2} \\right) = 45^{o} + \\dfrac{\\widehat{N}}{2}$ (2) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow$ $\\widehat{M_{1}} = \\widehat{M_{2}}$ hay $MP$ l\u00e0 tia ph\u00e2n gi\u00e1c $\\widehat{HMx}$ <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{MNH}$ c\u00f3 $K$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a m\u1ed9t tia ph\u00e2n gi\u00e1c ngo\u00e0i t\u1ea1i $M$ v\u00e0 tia ph\u00e2n gi\u00e1c trong g\u00f3c $N$ n\u00ean $HK$ l\u00e0 tia ph\u00e2n gi\u00e1c ngo\u00e0i t\u1ea1i \u0111\u1ec9nh $H$ do \u0111\u00f3 $\\widehat{KHP} = \\dfrac{\\widehat{MHP}}{2} = \\dfrac{90^{o}}{2} = 45^{o}$ <br\/> $\\blacktriangleright$ Ta c\u00f3 $\\widehat{N} = \\widehat{KHP} = 45^{o}$ hai g\u00f3c n\u00e0y \u1edf v\u1ecb tr\u00ed \u0111\u1ed3ng v\u1ecb n\u00ean $MN \/\/ HK$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: D <\/span> ","column":2}]}],"id_ques":1948},{"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 $ABC$, tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia $BC$ l\u1ea5y \u0111i\u1ec3m $M$ sao cho $MB = AB$. Tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia $CB$ l\u1ea5y \u0111i\u1ec3m $N$ sao cho $NC = AC$. Qua $M$ k\u1ebb \u0111\u01b0\u1eddng th\u1eb3ng song song v\u1edbi $AB$. Qua $N$ k\u1ebb \u0111\u01b0\u1eddng th\u1eb3ng song song v\u1edbi $AC$. Hai \u0111\u01b0\u1eddng th\u1eb3ng \u0111\u00f3 c\u1eaft nhau t\u1ea1i $P$. Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y sai? ","select":["A. $MA$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{PMB}$ ","B. $NA$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{PNC}$ ","C. $PA$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{MNP}$ ","D. C\u1ea3 $3$ \u0111\u00e1p \u00e1n tr\u00ean \u0111\u1ec1u \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> So s\u00e1nh $\\widehat{M_{1}}$ v\u00e0 $\\widehat{M_{2}}$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> So s\u00e1nh $\\widehat{N_{1}}$ v\u00e0 $\\widehat{N_{2}}$ <br\/> <b> B\u01b0\u1edbc 3: <\/b> D\u1ef1a v\u00e0o t\u00ednh ch\u1ea5t ba \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c trong tam gi\u00e1c \u0111\u1ec3 x\u00e9t \u0111\u00e1p \u00e1n 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/bai21/lv3/img\/H7C3B21_K08.png' \/><\/center> <br\/> <br\/> $\\blacktriangleright$ $\\triangle{AMB}$ c\u00f3 $AB = MB$ (gt) <br\/> $\\Rightarrow$ $\\triangle{AMB}$ c\u00e2n t\u1ea1i $B$ (\u0111\u1ecbnh ngh\u0129a), suy ra $\\widehat{A_{1}} = \\widehat{M_{1}}$ (t\u00ednh ch\u1ea5t tam gi\u00e1c c\u00e2n) <br\/> M\u1eb7t kh\u00e1c, $\\widehat{A_{1}} = \\widehat{M_{2}}$ (so le trong) <br\/> $\\Rightarrow$ $\\widehat{M_{1}} = \\widehat{M_{2}}$, n\u00ean $MA$ l\u00e0 tia ph\u00e2n gi\u00e1c $\\widehat{PMB}$ <br\/> $\\blacktriangleright$ $\\triangle{ACN}$ c\u00f3 $AC = CN$ (gt) <br\/> $\\Rightarrow$ $\\triangle{ACN}$ c\u00e2n t\u1ea1i $C$, suy ra $\\widehat{A_{2}} = \\widehat{N_{1}}$ (t\u00ednh ch\u1ea5t tam gi\u00e1c c\u00e2n) <br\/> M\u1eb7t kh\u00e1c, $\\widehat{A_{2}} = \\widehat{N_{2}}$ (so le trong) <br\/> $\\Rightarrow$ $\\widehat{N_{1}} = \\widehat{N_{2}}$ n\u00ean $NA$ l\u00e0 tia ph\u00e2n gi\u00e1c $\\widehat{PNC}$ <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{MNP}$ c\u00f3: <br\/> $MA$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{M}$ <br\/> $NA$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{N}$ <br\/> $\\Rightarrow$ $PA$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{MPN}$ (t\u00ednh ch\u1ea5t ba \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c trong tam gi\u00e1c) <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: D <\/span> ","column":2}]}],"id_ques":1949},{"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":[[["BC","CB","BD+DC","BD+CD","DB+CD","DB+DC"]]],"list":[{"point":10,"width":50,"content":"","type_input":"","ques":"Cho tam gi\u00e1c $ABC$ c\u00f3 $\\widehat{A}= 60^{o}$. Ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $B$ v\u00e0 g\u00f3c $C$ c\u1eaft c\u1ea1nh $AC$ v\u00e0 $AB$ l\u1ea7n l\u01b0\u1ee3t \u1edf $M$ v\u00e0 $N$. <br\/> Khi \u0111\u00f3: $BN + CM =$ _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 $BN = BD$ b\u1eb1ng c\u00e1ch ch\u1ee9ng minh $\\triangle{NIB} = \\triangle{DIB}$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> Ch\u1ee9ng minh $CM = CD$ b\u1eb1ng c\u00e1ch ch\u1ee9ng minh $\\triangle{MIC} = \\triangle{DIC}$ <br\/> <b> B\u01b0\u1edbc 3: <\/b> So s\u00e1nh $BN + CM$ v\u1edbi $BC$ <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai21/lv3/img\/H7C3B21_K09.png' \/><\/center> $\\blacktriangleright$ $\\triangle{ABC}$ c\u00f3 $\\widehat{A} = 60^{o}$ n\u00ean: <br\/> $\\widehat{B} + \\widehat{C} = 180^{o} - 60^{o} = 120^{o}$ (t\u1ed5ng ba g\u00f3c trong tam gi\u00e1c $ABC$) <br\/> $\\Rightarrow$ $\\widehat{B_{1}} + \\widehat{C_{1}} = \\dfrac{\\widehat{B} + \\widehat{C}}{2} = \\dfrac{120^{o}}{2} = 60^{o}$ (\u0111\u1ecbnh ngh\u0129a \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c) <br\/> $\\blacktriangleright$ G\u1ecdi $I$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $BM$ v\u00e0 $CN$ ta c\u00f3: <br\/> $\\widehat{BIC} = 180^{o} - (\\widehat{B_{1}} + \\widehat{C_{1}}) = 180^{o} - 60^{o} = 120^{o}$ (t\u1ed5ng ba g\u00f3c trong $\\triangle{BIC}$) <br\/> $\\Rightarrow$ $\\widehat{I_{1}} = \\widehat{I_{4}} = 180^{o} - \\widehat{BIC} = 180^{o} - 120^{o} = 60^{o}$ (c\u00f9ng k\u1ec1 b\u00f9 v\u1edbi $\\widehat{BIC}$) <br\/> K\u1ebb tia ph\u00e2n gi\u00e1c $ID$ c\u1ee7a $\\widehat{BIC}$, ta c\u00f3 $\\widehat{I_{2}} = \\widehat{I_{3}} = \\dfrac{\\widehat{BIC}}{2} = \\dfrac{120^{o}}{2} = 60^{o}$ <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{NIB}$ v\u00e0 $\\triangle{DIB}$ c\u00f3: <br\/> $\\begin{cases} \\widehat{B_{1}} = \\widehat{B_{2}} (gt) \\\\ BI \\hspace{0,2cm} \\text{chung} \\\\ \\widehat{I_{1}} = \\widehat{I_{2}} = 60^{o} (cmt) \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{NIB} = \\triangle{DIB}$ (g.c.g) n\u00ean $BN = BD$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) (1) <br\/> $\\blacktriangleright$ T\u01b0\u01a1ng t\u1ef1, $\\triangle{MIC} = \\triangle{DIC}$ (g.c.g) v\u00ec: <br\/> $\\begin{cases} \\widehat{C_{1}} = \\widehat{C_{2}} (gt) \\\\ IC \\hspace{0,2cm} \\text{chung} \\\\ \\widehat{I_{3}} = \\widehat{I_{4}} = 60^{o} (cmt) \\end{cases}$ <br\/> $\\Rightarrow$ $CM = CD$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) (2) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow$ $BN + CM = BD + CD = BC$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p c\u1ea7n \u0111i\u1ec1n l\u00e0: $BC$ <\/span>"}]}],"id_ques":1950}],"lesson":{"save":0,"level":3}}