{"segment":[{"time":24,"part":[{"title":"\u0110i\u1ec1n \u0111\u00e1p \u00e1n \u0111\u00fang v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["160"]]],"list":[{"point":10,"width":50,"type_input":"","ques":" Cho $\\widehat{xOy} = 80^{o}$, \u0111i\u1ec3m $A$ n\u1eb1m trong $\\widehat{xOy}$. V\u1ebd \u0111i\u1ec3m $B$ sao cho $Ox$ l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $AB$. V\u1ebd \u0111i\u1ec3m $C$ sao cho $Oy$ l\u00e0 trung tr\u1ef1c c\u1ee7a $AC$. T\u00ednh s\u1ed1 \u0111o g\u00f3c $BOC$. <br\/> <b> \u0110\u00e1p \u00e1n l\u00e0: <\/b> $\\widehat{BOC} = $ _input_ $^o$ ","hint":"Ch\u1ee9ng minh $O$ thu\u1ed9c \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $AB$ r\u1ed3i t\u00ednh $\\widehat{BOC}$","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 $O$ thu\u1ed9c \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $AB$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> T\u00ednh $\\widehat{BOC}$ d\u1ef1a v\u00e0o g\u00f3c $xOy$ <br\/><br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai22/lv3/img\/H7C3B22_K05.png' \/><\/center> <br\/> G\u1ecdi $I; K$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $AB$ v\u1edbi $Ox$ v\u00e0 $AC$ v\u1edbi $Oy$ <br\/> V\u00ec $Ox$ l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $AB$ (gt) <br\/> $\\Rightarrow$ $OA = OB$ (\u0111\u1ecbnh l\u00fd thu\u1eadn - t\u00ednh ch\u1ea5t \u0111\u01b0\u1eddng trung tr\u1ef1c) (1) <br\/> $Oy$ l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $AC$ (gt) <br\/> $\\Rightarrow$ $OA = OC$ (\u0111\u1ecbnh l\u00fd thu\u1eadn - t\u00ednh ch\u1ea5t \u0111\u01b0\u1eddng trung tr\u1ef1c) (2) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow$ $OB = OC$ $\\Rightarrow$ $O$ thu\u1ed9c \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $BC$ (\u0111\u1ecbnh l\u00fd \u0111\u1ea3o - t\u00ednh ch\u1ea5t \u0111\u01b0\u1eddng trung tr\u1ef1c) <br\/> $\\triangle{OAB}$ c\u00f3 $OA = OB$ n\u00ean $\\triangle{OAB}$ c\u00e2n t\u1ea1i $O$ <br\/> $\\Rightarrow$ $Ox$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{AOB}$ (T\u00ednh ch\u1ea5t trung tr\u1ef1c trong tam gi\u00e1c c\u00e2n) <br\/> $\\Rightarrow$ $ \\widehat{O_{1}} = \\widehat{O_{2}}$ <br\/> T\u01b0\u01a1ng t\u1ef1 ta c\u0169ng c\u00f3 $\\triangle{OAC}$ c\u00e2n t\u1ea1i $C$ $\\Rightarrow$ $\\widehat{O_{3}} = \\widehat{O_{4}}$ <br\/> $\\begin{align} \\widehat{BOC} &= \\widehat{O_{1}} + \\widehat{O_{2}} + \\widehat{O_{3}} + \\widehat{O_{4}} \\\\ &= \\widehat{O_{2}} + \\widehat{O_{2}} + \\widehat{O_{3}} + \\widehat{O_{3}} \\\\ &= 2\\widehat{O_{2}} + 2\\widehat{O_{3}} \\\\ &= 2(\\widehat{O_{2}} + \\widehat{O_{3}}) \\\\ &= 2 \\widehat{xOy} \\\\ &= 2 . 80^{o} \\\\ &= 160^{o} \\end{align}$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n l\u00e0: $160$ <\/span> <br\/> <span class='basic_green'> <i> Nh\u1eadn x\u00e9t: +) Trong m\u1ed9t tam gi\u00e1c c\u00e2n, \u0111\u01b0\u1eddng trung tr\u1ef1c \u1ee9ng v\u1edbi c\u1ea1nh \u0111\u00e1y \u0111\u1ed3ng th\u1eddi l\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c, \u0111\u01b0\u1eddng trung tuy\u1ebfn v\u00e0 \u0111\u01b0\u1eddng cao c\u00f9ng xu\u1ea5t ph\u00e1t t\u1eeb \u0111\u1ec9nh \u0111\u1ed1i di\u1ec7n v\u1edbi c\u1ea1nh \u0111\u00f3 <br\/> +) Trong tam gi\u00e1c \u0111\u1ec1u \u0111\u01b0\u1eddng cao c\u0169ng l\u00e0 \u0111\u01b0\u1eddng trung tuy\u1ebfn, \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c, \u0111\u01b0\u1eddng trung tr\u1ef1c <\/i> <\/span> "}]}],"id_ques":1971},{"time":24,"part":[{"title":"\u0110i\u1ec1n \u0111\u00e1p \u00e1n \u0111\u00fang v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["50"]]],"list":[{"point":10,"width":50,"type_input":"","ques":" Cho tam gi\u00e1c $ABC$ c\u00f3 $\\widehat{BAC} = 115^{o}$. C\u00e1c \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $AB$ v\u00e0 $AC$ l\u1ea7n l\u01b0\u1ee3t c\u1eaft c\u1ea1nh $BC$ \u1edf $Q$ v\u00e0 $N$. T\u00ednh s\u1ed1 \u0111o g\u00f3c $QAN$ <br\/> <b> \u0110\u00e1p \u00e1n l\u00e0: <\/b> $\\widehat{QAN} = $ _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 $\\widehat{B} = \\widehat{BAQ}$ <br\/> <b> B\u01b0\u1edbc 2: <\/b>Ch\u1ee9ng minh $\\widehat{C} = \\widehat{CAN}$ <br\/> <b> B\u01b0\u1edbc 3: <\/b> T\u00ednh $\\widehat{BAQ} + \\widehat{CAN}$ <br\/> <b> B\u01b0\u1edbc 4: <\/b> T\u00ednh $\\widehat{QAN}$ <br\/><br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai22/lv3/img\/H7C3B22_K14.png' \/><\/center> <br\/> G\u1ecdi $P$, $M$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AB; AC$ <br\/> Khi \u0111\u00f3, $PQ$ l\u00e0 trung tr\u1ef1c c\u1ee7a $AB$ $\\Rightarrow$ $QA = QB$ (t\u00ednh ch\u1ea5t) <br\/> $MN$ l\u00e0 trung tr\u1ef1c c\u1ee7a $AC$ $\\Rightarrow$ $NA = NC$ (t\u00ednh ch\u1ea5t) <br\/> $\\blacktriangleright$ X\u00e9t hai tam gi\u00e1c vu\u00f4ng $QAP$ v\u00e0 $QBP$ c\u00f3: <br\/> $\\begin{cases} PQ \\hspace{0,2cm} \\text{chung} \\\\ QA = QB (cmt) \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{QAP} = \\triangle{QBP}$ (c\u1ea1nh huy\u1ec1n, c\u1ea1nh g\u00f3c vu\u00f4ng) <br\/> $\\Rightarrow$ $\\widehat{B} = \\widehat{A_{1}}$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) (1) <br\/> Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1 ta c\u00f3: $\\triangle{NAM} = \\triangle{NCM}$ (c\u1ea1nh huy\u1ec1n - c\u1ea1nh g\u00f3c vu\u00f4ng) <br\/> $\\Rightarrow$ $\\widehat{A_{3}} = \\widehat{C}$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) (2) <br\/> $\\blacktriangleright$ $\\triangle{ABC}$ c\u00f3: $\\widehat{BAC} + \\widehat{B} + \\widehat{C} = 180^{o}$ (t\u1ed5ng ba g\u00f3c trong $1$ tam gi\u00e1c) <br\/> $ \\begin{align} \\Rightarrow \\widehat{B} + \\widehat{C} &= 180^{o} - \\widehat{BAC} \\\\ &= 180^{o} - 115^{o} \\\\ &= 65^{o} (3) \\end{align}$ <br\/> T\u1eeb (1), (2), (3) $\\Rightarrow$ $\\widehat{A_{1}} + \\widehat{A_{3}} = 65^{o}$ <br\/> $\\blacktriangleright$ M\u1eb7t kh\u00e1c ta c\u00f3: $\\widehat{A_{1}} + \\widehat{A_{2}} + \\widehat{A_{3}} = \\widehat{BAC}$ <br\/> $\\begin{align} \\Rightarrow \\widehat{A_{2}} &= \\widehat{BAC} - (\\widehat{A_{1}} + \\widehat{A_{3}}) \\\\ &= 115^{o} - 65^{o} \\\\ &= 50^{o} \\end{align}$ <br\/> Hay $\\widehat{QAN} = 50^{o}$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n l\u00e0: $50$ <\/span> "}]}],"id_ques":1972},{"time":24,"part":[{"title":"","title_trans":"Ch\u1ecdn \u0111\u01b0\u1ee3c nhi\u1ec1u \u0111\u00e1p \u00e1n","temp":"checkbox","correct":[["1","4"]],"list":[{"point":10,"img":"","ques":"H\u00e3y ch\u1ecdn nh\u1eefng \u0111\u00e1p \u00e1n \u0111\u00fang.","hint":"","column":1,"number_true":2,"select":["A. \u0110\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a m\u1ed9t \u0111o\u1ea1n th\u1eb3ng l\u00e0 t\u1eadp h\u1ee3p t\u1ea5t c\u1ea3 c\u00e1c \u0111i\u1ec3m c\u00e1ch \u0111\u1ec1u hai \u0111\u1ea7u m\u00fat c\u1ee7a \u0111o\u1ea1n th\u1eb3ng \u0111\u00f3. ","B. Trong tam gi\u00e1c c\u00e2n, tr\u1ecdng t\u00e2m v\u00e0 \u0111i\u1ec3m c\u00e1ch \u0111\u1ec1u ba \u0111\u1ec9nh tr\u00f9ng nhau","C. \u0110\u01b0\u1eddng tr\u00f2n ngo\u1ea1i ti\u1ebfp tam gi\u00e1c c\u00f3 t\u00e2m l\u00e0 giao \u0111i\u1ec3m ba ph\u00e2n gi\u00e1c c\u1ee7a tam gi\u00e1c \u0111\u00f3.","D. Trong tam gi\u00e1c \u0111\u1ec1u, ba \u0111\u01b0\u1eddng trung tr\u1ef1c c\u0169ng \u0111\u1ed3ng th\u1eddi l\u00e0 ba \u0111\u01b0\u1eddng trung tuy\u1ebfn, ba \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c."],"explain":" <span class='basic_left'> <b> A - \u0110\u00daNG <\/b> theo <b> Nh\u1eadn x\u00e9t <\/b> s\u00e1ch gi\u00e1o khoa To\u00e1n 7 - trang 75 <br\/> <b> B - SAI v\u00ec: <\/b> <br\/> V\u00ed d\u1ee5 nh\u01b0 h\u00ecnh v\u1ebd d\u01b0\u1edbi \u0111\u00e2y, $\\triangle{ABC}$ c\u00e2n t\u1ea1i $A$ nh\u01b0ng tr\u1ecdng t\u00e2m $G$ kh\u00e1c \u0111i\u1ec3m $I$ l\u00e0 \u0111i\u1ec3m c\u00e1ch \u0111\u1ec1u ba \u0111\u1ec9nh c\u1ee7a tam gi\u00e1c <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai22/lv3/img\/H7C3B22_K01.png' \/><\/center> <br\/> <b> C - Sai v\u00ec: <\/b> <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai22/lv3/img\/H7C3B22_K01C.png' \/><\/center> <br\/> $\\triangle{ABC}$ c\u00f3 $O$ l\u00e0 giao \u0111i\u1ec3m ba \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c n\u00ean ta c\u00f3: <br\/> $OH = OK = OM$ (t\u00ednh ch\u1ea5t) <br\/> $\\Rightarrow$ $O$ c\u00e1ch \u0111\u1ec1u ba c\u1ea1nh c\u1ee7a tam gi\u00e1c hay $O$ l\u00e0 t\u00e2m \u0111\u01b0\u1eddng tr\u00f2n n\u1ed9i ti\u1ebfp tam gi\u00e1c <br\/> <b> D - \u0110\u00daNG v\u00ec: <\/b> <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai22/lv3/img\/H7C3B22_K01D.png' \/><\/center> <br\/> Gi\u1ea3 s\u1eed $\\triangle{ABC}$ \u0111\u1ec1u c\u00f3 $AM; CN; BP$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a c\u1ea1nh $BC; AB$ v\u00e0 $AC$. Ta c\u1ea7n ch\u1ee9ng minh $AM, CN, BP$ c\u0169ng l\u00e0 \u0111\u01b0\u1eddng trung tuy\u1ebfn v\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c <br\/> $\\blacktriangleright$ $\\triangle{ABC}$ c\u00f3: $AM$ l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a c\u1ea1nh $BC$ <br\/> N\u00ean $BM = MC; AB = AC$ (t\u00ednh ch\u1ea5t) <br\/> V\u00ec $BM = MC$ n\u00ean $AM$ c\u0169ng l\u00e0 \u0111\u01b0\u1eddng trung tuy\u1ebfn \u1ee9ng v\u1edbi c\u1ea1nh $BC$ <br\/> X\u00e9t hai tam gi\u00e1c vu\u00f4ng $AMB$ v\u00e0 $AMC$ c\u00f3: <br\/> $\\begin{cases} AM \\hspace{0,2cm} \\text{chung} \\\\ AB = AC (\\triangle{ABC} \\hspace{0,2cm} \\text{\u0111\u1ec1u}) \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{AMB} = \\triangle{AMC}$ (c\u1ea1nh huy\u1ec1n - c\u1ea1nh g\u00f3c vu\u00f4ng) <br\/> $\\Rightarrow$ $\\widehat{BAM} = \\widehat{CAM}$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) <br\/> $\\Rightarrow$ $AM$ l\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{BAC}$ <br\/> $\\blacktriangleright$ Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1 ta c\u0169ng c\u00f3 $BN; CP$ c\u0169ng l\u00e0 \u0111\u1ed3ng th\u1eddi \u0111\u01b0\u1eddng trung tuy\u1ebfn v\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c <br\/><span class='basic_pink'> V\u1eady c\u00e1c \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: A; D <\/span> "}]}],"id_ques":1973},{"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 $ABC$ c\u00e2n t\u1ea1i $A$ c\u00f3 $\\widehat{B} = 36^{o}$. G\u1ecdi $O$ l\u00e0 giao \u0111i\u1ec3m ba \u0111\u01b0\u1eddng trung tr\u1ef1c, $I$ l\u00e0 giao \u0111i\u1ec3m ba \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c. Khi \u0111\u00f3 $BC$ l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a \u0111o\u1ea1n th\u1eb3ng $OI$. <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\u1ec9 ra $BC \\perp OI$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> T\u00ednh s\u1ed1 \u0111o g\u00f3c $A$ <br\/> <b> B\u01b0\u1edbc 3: <\/b> T\u00ednh s\u1ed1 \u0111o g\u00f3c $OBM$ <br\/> <b> B\u01b0\u1edbc 4: <\/b> So s\u00e1nh $MI$ v\u00e0 $MO$ b\u1eb1ng c\u00e1ch x\u00e9t hai tam gi\u00e1c $MBI$ v\u00e0 $MBO$ <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai22/lv3/img\/H7C3B22_K02.png' \/><\/center> <br\/> $\\blacktriangleright$ G\u1ecdi $M$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $BC$; $H$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AC$ <br\/> N\u1ed1i $B$ v\u1edbi $O$ <br\/> V\u00ec $OA$ l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $BC$ n\u00ean $OA \\perp BC$ t\u1ea1i $M$ (\u0111\u1ecbnh ngh\u0129a) <br\/> Hay $BC \\perp OI$ t\u1ea1i $M$ (1) <br\/> $\\blacktriangleright$ $\\triangle{ABC}$ c\u00e2n t\u1ea1i $A$ (gi\u1ea3 thi\u1ebft) n\u00ean $\\widehat{C} = \\widehat{B} = 36^{o}$ (t\u00ednh ch\u1ea5t) <br\/> $\\widehat{A} + \\widehat{B} + \\widehat{C} = 180^{o}$ (t\u1ed5ng ba g\u00f3c trong m\u1ed9t tam gi\u00e1c) <br\/> $\\begin{align} \\Rightarrow \\widehat{A} &= 180^{o} - (\\widehat{B} + \\widehat{C}) \\\\ &= 180^{o} - (36^{o} + 36^{o}) \\\\ &= 108^{o} \\end{align}$ <br\/> V\u00ec $AO$ l\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c g\u00f3c $A$ n\u00ean $\\widehat{A_{1}} = \\widehat{A_{2}} = \\dfrac{\\widehat{A}}{2} = \\dfrac{108^{o}}{2} = 54^{o}$ (\u0111\u1ecbnh ngh\u0129a) <br\/> $\\blacktriangleright$ M\u1eb7t kh\u00e1c do $O$ l\u00e0 giao \u0111i\u1ec3m ba \u0111\u01b0\u1eddng trung tr\u1ef1c (gi\u1ea3 thi\u1ebft) <br\/> N\u00ean $OA = OB$ (t\u00ednh ch\u1ea5t) <br\/> $\\triangle{AOB}$ c\u00f3 $OA = OB$ n\u00ean c\u00e2n t\u1ea1i $O$ (\u0111\u1ecbnh ngh\u0129a) <br\/> $\\Rightarrow$ $\\widehat{ABO} = \\widehat{A_{1}} = 54^{o}$ (t\u00ednh ch\u1ea5t) <br\/> $\\Rightarrow$ $\\widehat{B_{1}} = \\widehat{ABO} - \\widehat{ABC} = 54^{o} - 36^{o} = 18^{o}$ <br\/> $\\blacktriangleright$ X\u00e9t hai tam gi\u00e1c vu\u00f4ng $MBO$ v\u00e0 $MBI$ c\u00f3: <br\/> $\\begin{cases} BM \\hspace{0,2cm} \\text{chung} \\\\ \\widehat{MBI} = \\dfrac{1}{2}\\widehat{B} = \\dfrac{36^{o}}{2} = 18^{o} = \\widehat{B_{1}} \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{MBO} = \\triangle{MBI}$ (c\u1ea1nh g\u00f3c vu\u00f4ng - g\u00f3c nh\u1ecdn) <br\/> $\\Rightarrow$ $MI = MO$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) (2) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow$ $BC$ l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $OI$ <br\/> V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 <span class='basic_pink'> \u0110\u00daNG <\/span> ","column":2}]}],"id_ques":1974},{"time":24,"part":[{"title":"","title_trans":"Ch\u1ecdn \u0111\u01b0\u1ee3c nhi\u1ec1u \u0111\u00e1p \u00e1n","temp":"checkbox","correct":[["2","4","5","6"]],"list":[{"point":10,"img":"","ques":"Cho tam gi\u00e1c $ABC$ c\u00f3 $\\widehat{A} = 140^{o}$, c\u00e1c \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a c\u1ea1nh $AB$ v\u00e0 $AC$ c\u1eaft c\u1ea1nh $BC$ t\u1ea1i $E$ v\u00e0 $F$ v\u00e0 c\u1eaft nhau t\u1ea1i $I$. H\u00e3y ch\u1ecdn nh\u1eefng kh\u1eb3ng \u0111\u1ecbnh \u0111\u00fang. ","hint":"","column":2,"number_true":4,"select":["A. $\\triangle{BIC}$ \u0111\u1ec1u ","B. $\\triangle{ABE}$ c\u00e2n ","C. $\\widehat{BIC} = 90^{o}$ ","D. $\\widehat{BIC} = 80^{o}$ ","E. $ \\triangle{ACF}$ c\u00e2n ","F. $\\triangle{BIC}$ c\u00e2n "],"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 c\u00e1c c\u1eb7p c\u1ea1nh $EA$ v\u00e0 $EB$; $FA$ v\u00e0 $FC$ d\u1ef1a v\u00e0o t\u00ednh ch\u1ea5t \u0111\u01b0\u1eddng trung tr\u1ef1c \u0111\u1ec3 x\u00e9t \u0111\u00e1p \u00e1n $B, E$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> So s\u00e1nh $IB$ v\u00e0 $IC$ d\u1ef1a v\u00e0o t\u00ednh ch\u1ea5t \u0111\u01b0\u1eddng trung tr\u1ef1c \u0111\u1ec3 x\u00e9t \u0111\u00e1p \u00e1n $F$ <br\/> <b> B\u01b0\u1edbc 3: <\/b> T\u00ednh $\\widehat{BIC}$ \u0111\u1ec3 c\u00f3 k\u1ebft lu\u1eadn v\u1ec1 c\u00e1c \u0111\u00e1p \u00e1n $A, C, D$ <br\/><br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai22/lv3/img\/H7C3B22_K03.png' \/><\/center> <br\/> $\\blacktriangleright$ V\u00ec $E$ n\u1eb1m tr\u00ean \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $AB$ (gt) n\u00ean $EA = EB$ (t\u00ednh ch\u1ea5t) <br\/> $\\Rightarrow$ $\\triangle{ABE}$ c\u00e2n t\u1ea1i $E$ (\u0111\u1ecbnh ngh\u0129a) <br\/> $\\Rightarrow$ <b> \u0110\u00e1p \u00e1n B \u0111\u00fang <\/b> <br\/> V\u00ec $F$ thu\u1ed9c \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a c\u1ea1nh $AC$(gt) n\u00ean $AF = FC$ (t\u00ednh ch\u1ea5t) <br\/> $\\Rightarrow$ $\\triangle{ACF}$ c\u00e2n t\u1ea1i $F$ (\u0111\u1ecbnh ngh\u0129a) <br\/> $\\Rightarrow$ <b> \u0110\u00e1p \u00e1n E \u0111\u00fang <\/b> <br\/> $\\blacktriangleright$ Ta c\u00f3: $I$ n\u1eb1m tr\u00ean \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a c\u1ea1nh $AB$ n\u00ean $IA = IB$ (1) <br\/> $I$ n\u1eb1m tr\u00ean \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a c\u1ea1nh $AC$ n\u00ean $IA = IC$ (2) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow$ $IB = IC$, suy ra $\\triangle{BIC}$ c\u00e2n t\u1ea1i $I$ (\u0111\u1ecbnh ngh\u0129a) <br\/> $\\Rightarrow$ <b> \u0110\u00e1p \u00e1n F \u0111\u00fang <\/b> <br\/> $\\blacktriangleright$ T\u1eeb (1) $\\Rightarrow$ $\\triangle{AIB}$ c\u00e2n t\u1ea1i $I$ (\u0111\u1ecbnh ngh\u0129a) <br\/> $\\Rightarrow$ $\\widehat{IBA} = \\widehat{IAB}$ (3) <br\/> T\u1eeb (2) $\\Rightarrow$ $\\triangle{AIC}$ c\u00e2n t\u1ea1i $I$ (\u0111\u1ecbnh ngh\u0129a) <br\/> $\\Rightarrow$ $\\widehat{ICA} = \\widehat{IAC}$ (t\u00ednh ch\u1ea5t) (4) <br\/> M\u1eb7t kh\u00e1c ta c\u00f3: $\\widehat{IAC} + \\widehat{IAB} = \\widehat{BAC} = 140^{o}$ (5) <br\/> T\u1eeb (3), (4), (5) $\\Rightarrow$ $\\widehat{ICA} + \\widehat{IBA} = 140^{o}$ <br\/> T\u1ed5ng c\u00e1c g\u00f3c trong hai tam gi\u00e1c $AIB$ v\u00e0 $AIC$ l\u00e0 $360^{o}$, n\u00ean: <br\/> $ \\begin{align} \\widehat{IBA} + \\widehat{ICA} + \\widehat{BAC} + \\widehat{BIC} & = 360^{o}\\\\ 140^{o} + 140^{o} + \\widehat{BIC} &= 360^{o} \\\\ 280^{o} + \\widehat{BIC} &= 360^{o} \\\\ \\Rightarrow \\widehat{BIC} &= 360^{o} - 280^{o} \\\\ &= 80^{o} \\end{align}$ <br\/> $\\Rightarrow$ <b> \u0110\u00e1p \u00e1n A, C sai, \u0111\u00e1p \u00e1n D \u0111\u00fang <\/b> <br\/> <br\/><span class='basic_pink'> V\u1eady c\u00e1c \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: B; D; E; F <\/span> "}]}],"id_ques":1975},{"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 $ABC$ c\u00e2n t\u1ea1i $A$, c\u00e1c \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a c\u00e1c c\u1ea1nh $AB$ v\u00e0 $AC$ c\u1eaft \u0111\u01b0\u1eddng th\u1eb3ng $BC$ t\u1ea1i $N$ v\u00e0 $M$ ($N$ v\u00e0 $M$ n\u1eb1m ngo\u00e0i \u0111o\u1ea1n th\u1eb3ng $BC$). Tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia $AM$ l\u1ea5y \u0111i\u1ec3m $P$ sao cho $AP = MB$, khi \u0111\u00f3 $AM = AN = PC$. <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\/> T\u01b0\u01a1ng quan h\u00ecnh v\u1ebd ta c\u00f3 th\u1ec3 th\u1ea5y $AM = AN = PC$, t\u1eeb \u0111\u00f3 ta c\u1ea7n \u0111i ch\u1ee9ng minh <br\/> <b> B\u01b0\u1edbc 1: <\/b> Ch\u1ee9ng minh $\\triangle{MAK} = \\triangle{NAI}$ t\u1eeb \u0111\u00f3 suy ra $AM = AN$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> Ch\u1ee9ng minh $\\triangle{ABM} = \\triangle{CAP}$ \u0111\u1ec3 suy ra $AM = PC$ <br\/><br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span> <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai22/lv3/img\/H7C3B22_K04.png' \/><\/center> <br\/> G\u1ecdi $I; K$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AB$ v\u00e0 $AC$, n\u1ed1i $A$ v\u1edbi $N$, $P$ v\u1edbi $C$ <br\/> $\\blacktriangleright$ $M$ n\u1eb1m tr\u00ean \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $AC$ n\u00ean $MA = MC$ (t\u00ednh ch\u1ea5t) $\\Rightarrow$ $\\triangle{AMC}$ c\u00e2n t\u1ea1i $M$ (\u0111\u1ecbnh ngh\u0129a) <br\/> $N$ n\u1eb1m tr\u00ean \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $AB$ n\u00ean $NA = NB$ (t\u00ednh ch\u1ea5t) $\\Rightarrow$ $\\triangle{ANB}$ c\u00e2n t\u1ea1i $N$ (\u0111\u1ecbnh ngh\u0129a) <br\/> $\\blacktriangleright$ X\u00e9t hai tam gi\u00e1c vu\u00f4ng $AMK$ v\u00e0 $ANI$ c\u00f3: <br\/> $AI = \\dfrac{AB}{2} = \\dfrac{AC}{2} = AK$ (v\u00ec $AB = AC$)<br\/> $\\widehat{MAK} = \\widehat{C_{1}}$ ($\\triangle{AMC}$ c\u00e2n) <br\/> $\\widehat{BAN} = \\widehat{B_{1}}$ ($\\triangle{ANB}$ c\u00e2n) <br\/> M\u00e0 $\\widehat{B_{1}} = \\widehat{C_{1}}$ ($\\triangle{ABC}$ c\u00e2n t\u1ea1i $A$) $\\Rightarrow$ $\\widehat{MAK} = \\widehat{BAN}$ hay $\\widehat{MAK} = \\widehat{IAN}$<br\/> $\\Rightarrow$ $\\triangle{MAK} = \\triangle{ANI}$ (c\u1ea1nh g\u00f3c vu\u00f4ng - g\u00f3c nh\u1ecdn) <br\/> $\\Rightarrow$ $AM = AN$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) (1) <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{ABM}$ v\u00e0 $\\triangle{CAP}$ c\u00f3: <br\/> $AB = AC$ (gt); $BM = AP$ (gt) <br\/> $\\widehat{MBA} + \\widehat{ABC} = 180^{o}$ (2) <br\/> $\\widehat{PAC} + \\widehat{MAC} = 180^{o}$ (3) <br\/> M\u00e0 $\\widehat{ABN} = \\widehat{MCA}$ ($\\triangle{ABC}$ c\u00e2n) <br\/> V\u00e0 $\\widehat{CAM} = \\widehat{MCA}$ ($\\triangle{AMC}$ c\u00e2n) <br\/> $\\Rightarrow$ $\\widehat{ABN} = \\widehat{MAC}$ (4) <br\/> T\u1eeb (2), (3), (4) $\\Rightarrow$ $\\widehat{MBA} = \\widehat{PAC}$ <br\/> $\\Rightarrow$ $\\triangle{ABM} = \\triangle{CAP}$ (c.g.c) $\\Rightarrow$ $AM = PC$ (5) <br\/> T\u1eeb (1) v\u00e0 (5) $\\Rightarrow$ $AM = AN = PC$ <br\/> V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 <span class='basic_pink'> \u0110\u00daNG <\/span> ","column":2}]}],"id_ques":1976},{"time":4,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00fang hay sai","title_trans":"","temp":"true_false","correct":[["t","t","f","t"]],"list":[{"point":10,"image":"","col_name":["Cho tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i $A$. V\u1ebd ra ph\u00eda ngo\u00e0i tam gi\u00e1c $ABC$ c\u00e1c tam gi\u00e1c $ABD$ v\u00e0 $ACE$ l\u1ea7n l\u01b0\u1ee3t vu\u00f4ng c\u00e2n t\u1ea1i $D$ v\u00e0 $E$. G\u1ecdi $M$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a c\u1ea1nh $BC$, $F$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $MD$ v\u00e0 $AB$, $K$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $ME$ v\u00e0 $AC$ ","\u0110\u00fang","Sai"],"arr_ques":[" Ba \u0111i\u1ec3m $D, A, E$ th\u1eb3ng h\u00e0ng "," $DM \\perp AB$ v\u00e0 $EM \\perp AC$ "," $\\widehat{DME} = 85^{o}$ ","$ \\widehat{DME} = 90^{o}$"],"hint":"","explain":[" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai22/lv3/img\/H7C3B22_K06.png' \/><\/center> <br\/> \u0110\u00daNG v\u00ec: <br\/> $\\triangle{ACE}$ vu\u00f4ng c\u00e2n t\u1ea1i $E$ (gt) $\\Rightarrow$ $\\widehat{A_{1}} = 45^{o}$ <br\/> $\\triangle{ABD}$ vu\u00f4ng c\u00e2n t\u1ea1i $D$ (gt) $\\Rightarrow$ $\\widehat{A_{4}} = 45^{o}$ <br\/> $\\Rightarrow$ $\\widehat{EAD} = \\widehat{A_{1}} + \\widehat{CAB} + \\widehat{A_{4}} = 45^{o} + 90^{o} + 45^{o} = 180^{o}$ <br\/> $\\Rightarrow$ Ba \u0111i\u1ec3m $E, A, D$ th\u1eb3ng h\u00e0ng <\/span> "," <span class='basic_left'> <br\/> \u0110\u00daNG v\u00ec: <br\/> X\u00e9t $\\triangle{ABC}$ vu\u00f4ng t\u1ea1i $A$ c\u00f3 trung tuy\u1ebfn $AM$ \u1ee9ng v\u1edbi c\u1ea1nh huy\u1ec1n $BC$ <br\/> $\\Rightarrow$ $AM = MB = \\dfrac{1}{2}BC$ (t\u00ednh ch\u1ea5t trung tuy\u1ebfn trong tam gi\u00e1c vu\u00f4ng) <br\/> M\u1eb7t kh\u00e1c, ta c\u00f3: $DA = DB$ (v\u00ec $\\triangle{ADB}$ vu\u00f4ng c\u00e2n t\u1ea1i $D$) <br\/> $\\Rightarrow$ $M$ v\u00e0 $D$ thu\u1ed9c \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a c\u1ea1nh \u0111o\u1ea1n $AB$ <br\/> $\\Rightarrow$ $DM \\perp AB$ <br\/> Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1 ta c\u0169ng c\u00f3 $EM \\perp AC$ <\/span> "," <span class='basic_left'> <br\/> SAI v\u00ec: <br\/> X\u00e9t $\\triangle{ABD}$ vu\u00f4ng c\u00e2n t\u1ea1i $D$ c\u00f3 $DM$ l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $AB$ (cmt) <br\/> $\\Rightarrow$ $DM$ c\u0169ng l\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{ADB}$ (t\u00ednh ch\u1ea5t) <br\/> $\\Rightarrow$ $\\widehat{D_{1}} = 45^{o}$ <br\/> Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1 ta c\u0169ng c\u00f3 $\\widehat{E_{1}} = 45^{o}$ <br\/> $\\triangle{DEM}$ c\u00f3 $\\widehat{E_{1}} + \\widehat{D_{1}} + \\widehat{DME} = 180^{o}$ (t\u1ed5ng ba g\u00f3c trong 1 tam gi\u00e1c) <br\/> $\\Rightarrow$ $\\widehat{DME} = 180^{o} - 45^{o} - 45^{o} = 90^{o}$ <br\/> $\\Rightarrow$ $\\triangle{DEM}$ vu\u00f4ng c\u00e2n t\u1ea1i $M$ <\/span> "," <span class='basic_left'> <br\/> \u0110\u00daNG theo ch\u1ee9ng minh tr\u00ean <\/span> "]}]}],"id_ques":1977},{"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 $ABC$ c\u00f3 $\\widehat{A} = 100^{o}, \\widehat{C} = 30^{o}$. Tr\u00ean c\u1ea1nh $AC$ l\u1ea5y \u0111i\u1ec3m $D$ sao cho $\\widehat{CBD} = 10^{o}$. V\u1ebd \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{BAD}$ c\u1eaft c\u1ea1nh $BC$ t\u1ea1i $E$. Kh\u1eb3ng \u0111\u1ecbnh $AE$ kh\u00f4ng ph\u1ea3i l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $BD$. <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> T\u00ednh s\u1ed1 \u0111o g\u00f3c $ABC$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> T\u00ednh s\u1ed1 \u0111o g\u00f3c $ABD$ v\u00e0 $ADB$ <br\/> <b> B\u01b0\u1edbc 3: <\/b> X\u00e9t $\\triangle{AEB}$ v\u00e0 $\\triangle{AED}$ \u0111\u1ec3 so s\u00e1nh $EB$ v\u00e0 $ED$ t\u1eeb \u0111\u00f3 \u0111i \u0111\u1ebfn 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/bai22/lv3/img\/H7C3B22_K07.png' \/><\/center> <br\/> G\u1ecdi $I$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $AE$ v\u00e0 $BD$ <br\/> $\\blacktriangleright$ $\\triangle{ABC}$ c\u00f3 $\\widehat{A} = 100^{o}; \\widehat{C} = 30^{o}$ (gi\u1ea3 thi\u1ebft) <br\/> N\u00ean $\\widehat{ABC} = 180^{o} - (\\widehat{A} + \\widehat{C})$ (t\u1ed5ng ba g\u00f3c trong 1 tam gi\u00e1c b\u1eb1ng $180^{o}$) <br\/> $\\Rightarrow \\widehat{ABC} = 180^{o} - (100^{o} + 30^{o}) = 50^{o}$ <br\/> L\u1ea1i c\u00f3 $\\widehat{CBD} = 10^{o}$ (gi\u1ea3 thi\u1ebft) n\u00ean $\\widehat{ABD} = \\widehat{ABC} - \\widehat{CBD} = 50^{o} - 10^{o} = 40^{o}$ <br\/> M\u1eb7t kh\u00e1c $\\widehat{ADB}$ l\u00e0 g\u00f3c ngo\u00e0i t\u1ea1i \u0111\u1ec9nh $D$ c\u1ee7a $\\triangle{BDC}$ <br\/> N\u00ean $\\widehat{ADB} = \\widehat{CBD} + \\widehat{C} = 10^{o} + 30^{o} = 40^{o}$ <br\/> $\\Rightarrow$ $\\widehat{ABD} = \\widehat{ADB} = 40^{o}$, do \u0111\u00f3 $\\triangle{ABD}$ c\u00e2n t\u1ea1i $A$ <br\/> $\\Rightarrow$ $AB = AD$ (\u0111\u1ecbnh ngh\u0129a) <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{AEB}$ v\u00e0 $\\triangle{AED}$ c\u00f3: <br\/> $\\begin{cases} AB = AD \\\\ \\widehat{EAB} = \\widehat{EAD} (gt) \\\\ AE \\hspace{0,2cm} \\text{chung} \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{AEB} = \\triangle{AED}$ (c.g.c), suy ra $EB = ED$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) <br\/> $\\blacktriangleright$ Ta c\u00f3: $AB = AD$ n\u00ean $A$ thu\u1ed9c \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $BD$ (t\u00ednh ch\u1ea5t) (1) <br\/> $EB = ED$ n\u00ean $E$ thu\u1ed9c \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $BD$ (t\u00ednh ch\u1ea5t) (2) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow$ $AE$ l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a \u0111o\u1ea1n th\u1eb3ng $BD$ <br\/> V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 <span class='basic_pink'> SAI <\/span> <br\/> <span class='basic_green'> <i> Nh\u1eadn x\u00e9t: C\u00f3 th\u1ec3 s\u1eed d\u1ee5ng t\u00ednh ch\u1ea5t \u0111\u01b0\u1eddng trung tuy\u1ebfn \u1ee9ng v\u1edbi c\u1ea1nh \u0111\u00e1y \u0111\u1ed3ng th\u1eddi l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a tam gi\u00e1c c\u00e2n \u0111\u1ec3 l\u1eadp lu\u1eadn nh\u01b0 sau: <br\/> X\u00e9t $\\triangle{AIB}$ v\u00e0 $\\triangle{AID}$ c\u00f3: <br\/> $\\begin{cases} AB = AD \\\\ \\widehat{IAB} = \\widehat{IAD} (gt) \\\\ AI \\hspace{0,2cm} \\text{chung} \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{AIB} = \\triangle{AID}$ (c.g.c), suy ra $IB = ID$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) <br\/> $\\triangle{ABD}$ c\u00e2n t\u1ea1i $A$ c\u00f3 $AI$ l\u00e0 \u0111\u01b0\u1eddng trung tuy\u1ebfn \u1ee9ng v\u1edbi c\u1ea1nh \u0111\u00e1y $BD$ n\u00ean $AI$ l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a \u0111o\u1ea1n th\u1eb3ng $BD$. <br\/> Suy ra $AE$ l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a \u0111o\u1ea1n th\u1eb3ng $BD$ <\/i> <\/span> ","column":2}]}],"id_ques":1978},{"time":24,"part":[{"title":"\u0110i\u1ec1n \u0111\u00e1p \u00e1n \u0111\u00fang v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["90"]]],"list":[{"point":10,"width":50,"type_input":"","ques":" Cho tam gi\u00e1c nh\u1ecdn $ABC$ c\u00f3 $\\widehat{A} = 45^{o}$. C\u00e1c \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $AB, AC$ c\u1eaft nhau \u1edf $I$ v\u00e0 c\u1eaft c\u1ea1nh $BC$ l\u1ea7n l\u01b0\u1ee3t t\u1ea1i $E$ v\u00e0 $F$. T\u00ednh s\u1ed1 \u0111o g\u00f3c $EAF$. <br\/> <b> \u0110\u00e1p \u00e1n l\u00e0: <\/b> $\\widehat{EAF} = $ _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 s\u1ed1 \u0111o $\\widehat{EIF}$ th\u00f4ng qua t\u00ednh s\u1ed1 \u0111o $\\widehat{MIN}$ ($M$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AC$, $N$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AB$) <br\/> <b> B\u01b0\u1edbc 2: <\/b> T\u00ednh $\\widehat{IEF} + \\widehat{IFE}$ <br\/> <b> B\u01b0\u1edbc 3: <\/b> T\u00ednh $\\widehat{AFE} + \\widehat{AEF}$ th\u00f4ng qua vi\u1ec7c ch\u1ee9ng minh $FM, EN$ l\u00e0 c\u00e1c tia ph\u00e2n gi\u00e1c <br\/> <b> B\u01b0\u1edbc 4: <\/b> T\u00ednh $\\widehat{EAF}$ <br\/><br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai22/lv3/img\/H7C3B22_K08.png' \/><\/center> <br\/> G\u1ecdi $M; N$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 trung \u0111i\u1ec3m c\u1ee7a c\u00e1c c\u1ea1nh $AC$ v\u00e0 $AB$ <br\/> $\\blacktriangleright$ X\u00e9t hai tam gi\u00e1c vu\u00f4ng $AIM$ v\u00e0 $AIN$ c\u00f3: <br\/> $\\widehat{NAI} + \\widehat{AIN} + \\widehat{INA} = 180^{o}$ (t\u1ed5ng ba g\u00f3c trong $\\triangle{AIN}$) (1) <br\/> $\\widehat{AMI} + \\widehat{MIA} + \\widehat{IAM} = 180^{o}$ (t\u1ed5ng ba g\u00f3c trong $\\triangle{AMI}$) (2) <br\/> C\u1ed9ng theo v\u1ebf c\u1ee7a (1) v\u1edbi (2), ta \u0111\u01b0\u1ee3c: <br\/> $ \\begin{align} \\widehat{NAI} + \\widehat{AIN} + \\widehat{INA} + \\widehat{AMI} + \\widehat{MIA} + \\widehat{IAM} &= 180^{o} + 180^{o} \\\\ (\\widehat{NAI} + \\widehat{IAM}) + (\\widehat{AIN} + \\widehat{MIA}) + 90^{o} + 90^{o} &= 360^{o} \\\\ \\widehat{MAN} + \\widehat{MIN} &= 180^{o} \\\\ \\widehat{MIN} &= 180^{o} - \\widehat{MAN} \\\\ &= 180^{o} - 45^{o} \\\\ &= 135^{o} \\end{align}$ <br\/> $\\Rightarrow$ $\\widehat{EIF} = \\widehat{MIN} = 135^{o}$ (\u0111\u1ed1i \u0111\u1ec9nh) <br\/> $\\blacktriangleright$ $\\triangle{EIF}$ c\u00f3: $\\widehat{E_{1}} + \\widehat{F_{1}} + \\widehat{EIF} = 180^{o}$ (t\u1ed5ng ba g\u00f3c trong m\u1ed9t tam gi\u00e1c) <br\/> $\\Rightarrow$ $\\widehat{E_{1}} + \\widehat{F_{1}} = 180^{o} - \\widehat{EIF} = 180^{o} - 135^{o} = 45^{o}$ (1) <br\/> $\\blacktriangleright$ V\u00ec $F$ thu\u1ed9c \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $AC$ n\u00ean $FA = FC$, suy ra $\\triangle{FAC}$ c\u00e2n t\u1ea1i $F$ (\u0111\u1ecbnh ngh\u0129a) <br\/> $\\triangle{FAC}$ c\u00e2n t\u1ea1i $F$ c\u00f3 $FM$ l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c n\u00ean $FM$ c\u0169ng l\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c (t\u00ednh ch\u1ea5t) <br\/> $\\Rightarrow$ $\\widehat{F_{1}} = \\widehat{F_{2}}$ (\u0111\u1ecbnh ngh\u0129a) (2) <br\/> Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1 ta c\u00f3 $\\widehat{E_{1}} = \\widehat{E_{2}}$ (3) <br\/> T\u1eeb (1), (2) v\u00e0 (3) $\\Rightarrow$ $\\widehat{AFE} + \\widehat{AEF} = 2 .(\\widehat{E_{1}} + \\widehat{F_{1}}) = 2 . 45^{o} = 90^{o}$ <br\/> Do \u0111\u00f3 $\\widehat{EAF} = 90^{o}$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n l\u00e0: $90$ <\/span> "}]}],"id_ques":1979},{"time":24,"part":[{"title":"H\u00e3y s\u1eafp x\u1ebfp c\u00e1c \u00fd sau \u0111\u1ec3 \u0111\u01b0\u1ee3c th\u1ee9 t\u1ef1 ch\u1ee9ng minh \u0111\u00fang","title_trans":" Cho tam gi\u00e1c $ABC$. Hai \u0111i\u1ec3m $M; N$ theo th\u1ee9 t\u1ef1 di chuy\u1ec3n tr\u00ean hai tia $BA$ v\u00e0 $CA$ sao cho $BM + CN = BC$. Ch\u1ee9ng minh c\u00e1c \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $MN$ lu\u00f4n \u0111i qua m\u1ed9t \u0111i\u1ec3m c\u1ed1 \u0111\u1ecbnh.","temp":"sequence","correct":[[[2],[4],[3],[5],[1]]],"list":[{"point":10,"image":"","left":[" X\u00e9t $\\triangle{OMB}$ v\u00e0 $\\triangle{ODB}$ c\u00f3: <br\/> $OB$ chung; $\\widehat{B_{1}} = \\widehat{B_{2}}$ (c\u00e1ch v\u1ebd); $BD = BM$ (c\u00e1ch v\u1ebd) <br\/> $\\Rightarrow$ $\\triangle{OMB} = \\triangle{ODB}$ (c.g.c) "," Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1, ta \u0111\u01b0\u1ee3c $\\triangle{OCN} = \\triangle{OCD}$ (c.g.c) <br\/> $\\Rightarrow$ $ON = OD$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) (2) "," $\\Rightarrow$ $OM = OD$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) (1) "," T\u1eeb (1) v\u00e0 (2) $\\Rightarrow$ $OM = ON$, suy ra $O$ thu\u1ed9c \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $MN$ (t\u00ednh ch\u1ea5t) <br\/> $\\Rightarrow$ \u0110\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $MN$ lu\u00f4n \u0111i qua \u0111i\u1ec3m $O$ c\u1ed1 \u0111\u1ecbnh ","V\u1ebd tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $B$ v\u00e0 g\u00f3c $C$, ch\u00fang c\u1eaft nhau t\u1ea1i \u0111i\u1ec3m $O$ th\u00ec \u0111i\u1ec3m $O$ c\u1ed1 \u0111\u1ecbnh. <br\/> Tr\u00ean c\u1ea1nh $BC$ l\u1ea5y \u0111i\u1ec3m $D$ sao cho $BD = BM; CD = CN$ "],"top":100,"hint":"","explain":"<span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai22/lv3/img\/H7C3B22_K09.png' \/><\/center> <br\/> $\\blacktriangleright$ V\u1ebd tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $B$ v\u00e0 g\u00f3c $C$, ch\u00fang c\u1eaft nhau t\u1ea1i \u0111i\u1ec3m $O$ th\u00ec \u0111i\u1ec3m $O$ c\u1ed1 \u0111\u1ecbnh. <br\/> Tr\u00ean c\u1ea1nh $BC$ l\u1ea5y \u0111i\u1ec3m $D$ sao cho $BD = BM; CD = CN$ <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{OMB}$ v\u00e0 $\\triangle{ODB}$ c\u00f3: <br\/> $ \\left. \\begin{array}{l} OB \\hspace{0,2cm} \\text{chung} \\\\ \\widehat{B_{1}} = \\widehat{B_{2}} (\\text{c\u00e1ch} \\hspace{0,2cm} \\text{v\u1ebd}) \\\\ BD = BM (\\text{c\u00e1ch} \\hspace{0,2cm} \\text{v\u1ebd})\\end{array} \\right\\}$ $\\Rightarrow$ $\\triangle{OMB} = \\triangle{ODB}$ (c.g.c) <br\/> $\\Rightarrow$ $OM = OD$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) (1) <br\/> $\\blacktriangleright$ Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1, ta \u0111\u01b0\u1ee3c $\\triangle{OCN} = \\triangle{OCD}$ (c.g.c) <br\/> $\\Rightarrow$ $ON = OD$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) (2) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow$ $OM = ON$, suy ra $O$ thu\u1ed9c \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $MN$ (t\u00ednh ch\u1ea5t) <br\/> $\\Rightarrow$ \u0110\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $MN$ lu\u00f4n \u0111i qua \u0111i\u1ec3m $O$ c\u1ed1 \u0111\u1ecbnh "}]}],"id_ques":1980}],"lesson":{"save":0,"level":3}}