{"segment":[{"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 $MIK$ c\u00f3 $MI < MK$. G\u1ecdi $H$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $IK$. <br\/> Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y \u0111\u00fang? ","select":["A. $\\widehat{IMH} = \\widehat{HMK}$ ","B. $\\widehat{IMH} > \\widehat{HMK}$","C. $\\widehat{IMH} < \\widehat{HMK}$"],"hint":"V\u1ebd th\u00eam \u0111i\u1ec3m $D$ sao cho $H$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $MD$","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> V\u1ebd th\u00eam \u0111i\u1ec3m $D$ sao cho $H$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $MD$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> So s\u00e1nh $\\widehat{M_{1}}$ v\u00e0 $\\widehat{D}$ b\u1eb1ng c\u00e1ch ch\u1ee9ng minh $\\triangle{IHM} = \\triangle{KHD}$ <br\/> <b> B\u01b0\u1edbc 3: <\/b> So s\u00e1nh $\\widehat{D}$ v\u00e0 $\\widehat{M_{2}}$ b\u1eb1ng c\u00e1ch so s\u00e1nh $MK$ v\u00e0 $KD$ <br\/> <b> B\u01b0\u1edbc 4: <\/b> So s\u00e1nh $\\widehat{M_{1}}$ v\u00e0 $\\widehat{M_{2}}$ <br\/><br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai17/lv3/img\/H7C3B17_TB01.png' \/><\/center> <br\/> V\u1ebd \u0111i\u1ec3m $D$ sao cho $H$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $MD$ <br\/> X\u00e9t $\\triangle{IHM} = \\triangle{KHD}$ c\u00f3: <br\/> $\\begin{cases} IH = HK (gt) \\\\ \\widehat{IHM} = \\widehat{KHD} (\\text{\u0111\u1ed1i} \\hspace{0,2cm} \\text{\u0111\u1ec9nh}) \\\\ MH = HD (\\text{c\u00e1ch} \\hspace{0,2cm} \\text{v\u1ebd}) \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{IHM} = \\triangle{KHD}$ (c. g. c) <br\/> $\\Rightarrow$ $\\widehat{M_{1}} = \\widehat{D}$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) (1) <br\/> V\u00e0 $IM = KD$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) (2) <br\/> M\u1eb7t kh\u00e1c ta c\u00f3: $MI < MK$ (gt) (3) <br\/> T\u1eeb (2) v\u00e0 (3) $\\Rightarrow$ $KD < MK$ <br\/> $\\Rightarrow$ $\\widehat{M_{2}} < \\widehat{D}$ (quan h\u1ec7 gi\u1eefa g\u00f3c v\u00e0 c\u1ea1nh \u0111\u1ed1i di\u1ec7n trong tam gi\u00e1c $KMD$) (4) <br\/> T\u1eeb (1) v\u00e0 (4) $\\Rightarrow \\widehat{M_{2}} < \\widehat{M_{1}}$ hay $\\widehat{IMH} > \\widehat{HMK}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: B <\/span> ","column":3}]}],"id_ques":1821},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"Cho tam gi\u00e1c $MNP$ vu\u00f4ng t\u1ea1i $M$, tia ph\u00e2n gi\u00e1c g\u00f3c $N$ c\u1eaft $MP$ \u1edf $I$. <br\/> Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y \u0111\u00fang? ","select":["A. $MI = IP$ ","B. $MI > IP$","C. $MI < IP$"],"hint":"K\u1ebb $IH$ vu\u00f4ng g\u00f3c v\u1edbi $NP$ ","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> K\u1ebb $IH$ vu\u00f4ng g\u00f3c v\u00f3i $NP$ sau \u0111\u00f3 so s\u00e1nh $IM$ v\u00e0 $IH$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> So s\u00e1nh $IH$ v\u00e0 $IP$ <br\/> <b> B\u01b0\u1edbc 3: <\/b> So s\u00e1nh $IM$ v\u00e0 $IP$ <br\/><br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai17/lv3/img\/H7C3B17_TB02.png' \/><\/center> <br\/>K\u1ebb $IH$ vu\u00f4ng g\u00f3c v\u1edbi $NP$ <br\/> X\u00e9t $\\triangle{NMI}$ v\u00e0 $\\triangle{NHI}$ c\u00f3: <br\/> $\\begin{cases} \\widehat{NMI} = \\widehat{NHI} = 90^{o} \\\\ \\text{C\u1ea1nh} \\hspace{0,2cm} IN \\hspace{0,2cm} \\text{chung} \\\\ \\widehat{MNI} = \\widehat{HNI} (gt) \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{NMI} = \\triangle{NHI}$ (g.c.g) <br\/> $\\Rightarrow$ $IM = IH$ (c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) (1) <br\/> Trong $\\triangle{IHP}$ c\u00f3: <br\/> $\\widehat{IHP} = 90^{o}$ (c\u00e1ch l\u1ea5y \u0111i\u1ec3m $H$) <br\/> $\\Rightarrow$ $IH < IP$ (trong tam gi\u00e1c vu\u00f4ng c\u1ea1nh huy\u1ec1n l\u00e0 c\u1ea1nh l\u1edbn nh\u1ea5t) (2) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow$ $IM < IP$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: C <\/span> <br\/> <span class='basic_green'> <i> Nh\u1eadn x\u00e9t: \u1ede b\u00e0i n\u00e0y ta ph\u1ea3i so s\u00e1nh $MI$ v\u00e0 $IP$ kh\u00f4ng ph\u1ea3i l\u00e0 hai c\u1ea1nh c\u1ee7a m\u1ed9t tam gi\u00e1c n\u00ean kh\u00f4ng v\u1eadn d\u1ee5ng \u0111\u01b0\u1ee3c \u0111\u1ecbnh l\u00fd v\u1ec1 quan h\u1ec7 gi\u1eefa g\u00f3c v\u00e0 c\u1ea1nh \u0111\u1ed1i di\u1ec7n trong m\u1ed9t tam gi\u00e1c. Ta chuy\u1ec3n $MI$ v\u00e0 $IP$ v\u1ec1 c\u00f9ng m\u1ed9t tam gi\u00e1c b\u1eb1ng c\u00e1ch v\u1ebd th\u00eam \u0111\u01b0\u1eddng ph\u1ee5 $IH$. L\u00fac \u0111\u00f3 $MI = IH$, ta ch\u1ec9 c\u00f2n ph\u1ea3i so s\u00e1nh $MI$ v\u00e0 $IP$ \u1edf c\u00f9ng m\u1ed9t tam gi\u00e1c <\/i> <\/span> ","column":3}]}],"id_ques":1822},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"Cho tam gi\u00e1c $KHL$ c\u00f3 $KH < KL$, tia ph\u00e2n gi\u00e1c g\u00f3c $K$ c\u1eaft $HL$ \u1edf $D$. <br\/> Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y \u0111\u00fang? ","select":["A. $HD = DL$ ","B. $HD > DL$","C. $HD < DL$"],"hint":"Tr\u00ean $HL$ l\u1ea5y \u0111i\u1ec3m $I$ sao cho $KH = KI$ ","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> Tr\u00ean $KL$ l\u1ea5y \u0111i\u1ec3m $I$ sao cho $KH = KI$ <br\/> Sau \u0111\u00f3 so s\u00e1nh $HD$ v\u00e0 $ID$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> So s\u00e1nh $\\widehat{H_{1}}$ v\u00e0 $\\widehat{I_{1}}$ <br\/> <b> B\u01b0\u1edbc 3: <\/b> So s\u00e1nh $\\widehat{H_{1}}$ v\u00e0 $\\widehat{L}$ <br\/> <b> B\u01b0\u1edbc 4: <\/b> So s\u00e1nh $\\widehat{I_{1}}$ v\u00e0 $\\widehat{L}$ t\u1eeb \u0111\u00f3 so s\u00e1nh \u0111\u01b0\u1ee3c $HD$ v\u00e0 $DL$ <br\/><br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai17/lv3/img\/H7C3B17_TB03.png' \/><\/center> <br\/> Tr\u00ean $KL$ l\u1ea5y \u0111i\u1ec3m $I$ sao cho $KH = KI$ <br\/> X\u00e9t $\\triangle{KHD}$ v\u00e0 $\\triangle{KID}$ c\u00f3: <br\/> $\\begin{cases} KH = KI (\\text{c\u00e1ch} \\hspace{0,2cm} \\text{l\u1ea5y} \\hspace{0,2cm} \\text{\u0111i\u1ec3m} \\hspace{0,2cm} I) \\\\ \\widehat{HKD} = \\widehat{IKD} (gt) \\\\ \\text{C\u1ea1nh} \\hspace{0,2cm} KD \\hspace{0,2cm} \\text{chung} \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{HKD} = \\triangle{IKD}$ (c. g . c) <br\/> $\\Rightarrow$ $HD = ID$ (c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) (1) <br\/> Ta c\u00f3: $\\widehat{H_{1}}$ l\u00e0 g\u00f3c ngo\u00e0i t\u1ea1i $H$ c\u1ee7a tam gi\u00e1c $HKD$ n\u00ean: $\\widehat{H_{1}} = \\widehat{HDK} + \\widehat{HKD}$ <br\/> $\\widehat{I_{1}}$ l\u00e0 g\u00f3c ngo\u00e0i t\u1ea1i $I$ c\u1ee7a tam gi\u00e1c $KID$ n\u00ean: $\\widehat{I_{1}} = \\widehat{IDK} + \\widehat{IKD}$ <br\/> M\u00e0 $\\widehat{HDK} = \\widehat{IDK}$ (v\u00ec $\\triangle{HDK} = \\triangle{IDK}$) <br\/> $\\widehat{HKD} = \\widehat{IKD}$ (gt) <br\/> N\u00ean: $\\widehat{H_{1}} = \\widehat{I_{1}}$ (2) <br\/> M\u1eb7t kh\u00e1c: $\\widehat{H_{1}}$ l\u00e0 g\u00f3c ngo\u00e0i t\u1ea1i $H$ c\u1ee7a tam gi\u00e1c $HKL$ n\u00ean $\\widehat{H_{1}} > \\widehat{L}$ (3) <br\/> T\u1eeb (2) v\u00e0 (3) $\\Rightarrow$ $\\widehat{I_{1}} > \\widehat{L}$ <br\/> X\u00e9t $\\triangle{DIL}$ c\u00f3 $\\widehat{I_{1}} >\\widehat{L}$ <br\/> N\u00ean $DL > DI$ (\u0111\u1ecbnh l\u00fd quan h\u1ec7 gi\u1eefa g\u00f3c v\u00e0 c\u1ea1nh \u0111\u1ed1i di\u1ec7n trong tam gi\u00e1c) (4) <br\/> T\u1eeb (1) v\u00e0 (4) $\\Rightarrow$ $HD < DL$ <br\/> <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: C <\/span> <br\/> <span class='basic_green'> <i> Nh\u1eadn x\u00e9t: +) Sai l\u1ea7m c\u00f3 th\u1ec3 m\u1eafc ph\u1ea3i l\u00e0 t\u01b0\u1edfng r\u1eb1ng $\\widehat{HKD} = \\widehat{DKL}$ th\u00ec $HD = DL$ <br\/> +) L\u01b0u \u00fd r\u1eb1ng \u0111\u1ec3 so s\u00e1nh hai \u0111o\u1ea1n th\u1eb3ng n\u00ean \u0111\u01b0a v\u1ec1 c\u1ea1nh c\u1ee7a m\u1ed9t tam gi\u00e1c \u0111\u1ec3 c\u00f3 th\u1ec3 \u00e1p d\u1ee5ng \u0111\u1ecbnh l\u00fd v\u1ec1 quan h\u1ec7 gi\u1eefa g\u00f3c v\u00e0 c\u1ea1nh \u0111\u1ed1i di\u1ec7n trong m\u1ed9t tam gi\u00e1c. <br\/> +) Ngo\u00e0i ra \u1edf b\u00e0i n\u00e0y c\u00f2n s\u1eed d\u1ee5ng quan h\u1ec7 gi\u1eefa g\u00f3c ngo\u00e0i c\u1ee7a tam gi\u00e1c v\u1edbi g\u00f3c trong kh\u00f4ng k\u1ec1 \u0111\u1ec3 so s\u00e1nh g\u00f3c <\/i> <\/span> ","column":3}]}],"id_ques":1823},{"time":24,"part":[{"title":"\u0110i\u1ec1n \u0111\u00e1p \u00e1n \u0111\u00fang v\u00e0o \u00f4 tr\u1ed1ng ","title_trans":"","temp":"fill_the_blank","correct":[[["NP","PN"]]],"list":[{"point":10,"width":50,"content":"","type_input":"","ques":"Cho tam gi\u00e1c $MNP$. C\u00e1c tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $N$ v\u00e0 g\u00f3c $P$ c\u1eaft nhau t\u1ea1i $O$. <br\/> Khi \u0111\u00f3 trong tam gi\u00e1c $NOP$ c\u1ea1nh l\u1edbn nh\u1ea5t l\u00e0 c\u1ea1nh: _input_ ","hint":"T\u00ednh s\u1ed1 \u0111o $\\widehat{NOP}$ r\u1ed3i so s\u00e1nh","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> \u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd t\u1ed5ng ba g\u00f3c cho $\\triangle{MNP}$ v\u00e0 $\\triangle{NOP}$ \u0111\u1ec3 t\u00ednh s\u1ed1 \u0111o $\\widehat{NOP}$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> So s\u00e1nh $\\widehat{NOP}$ v\u1edbi $90^{o}$ \u0111\u1ec3 t\u00ecm c\u1ea1nh l\u1edbn nh\u1ea5t trong $\\triangle{NOP}$ <br\/><br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai17/lv3/img\/H7C3B17_TB11.png' \/><\/center> <br\/> $\\triangle{MNP}$ c\u00f3 $NO$ v\u00e0 $PO$ l\u00e0 c\u00e1c tia ph\u00e2n gi\u00e1c n\u00ean: <br\/> $\\widehat{N_{1}} = \\widehat{N_{2}} = \\widehat{\\dfrac{N}{2}}$; $\\widehat{P_{1}} = \\widehat{P_{2}} = \\widehat{\\dfrac{P}{2}}$ <br\/> N\u00ean $\\widehat{N_{1}} + \\widehat{P_{1}} = \\dfrac{1}{2}(\\widehat{N} + \\widehat{P})$ (1) <br\/> $\\triangle{NOP}$ c\u00f3: $\\widehat{N_{1}} + \\widehat{NOP} + \\widehat{P_{1}} = 180^{o}$ (t\u1ed5ng 3 g\u00f3c trong tam gi\u00e1c) <br\/> $\\Rightarrow$ $\\widehat{NOP} = 180^{o} - (\\widehat{N_{1}} + \\widehat{P_{1}})$ (2) <br\/> M\u1eb7t kh\u00e1c ta c\u00f3: $\\widehat{M} + \\widehat{N} + \\widehat{P} = 180^{o}$ (t\u1ed5ng 3 g\u00f3c trong tam gi\u00e1c) <br\/> $\\Rightarrow$ $\\widehat{N} + \\widehat{P} = 180^{o} - \\widehat{M}$ (3) <br\/> Thay (1) v\u00e0o (2), ta c\u00f3: $\\widehat{NOP} = 180^{o} - \\dfrac{1}{2}(\\widehat{N} + \\widehat{P})$ (4) <br\/> Thay (3) v\u00e0o (4), ta c\u00f3: $\\widehat{NOP} = 180^{o} - \\dfrac{1}{2}( 180^{o} - \\widehat{M}) = 90^{o} + \\dfrac{1}{2}\\widehat{M}$ <br\/> $\\Rightarrow$ $\\widehat{NOP} > 90^{o}$ hay $\\widehat{NOP}$ l\u00e0 g\u00f3c t\u00f9 <br\/> Do \u0111\u00f3 trong $\\triangle{NOP}$ c\u00f3 $NP$ l\u00e0 c\u1ea1nh l\u1edbn nh\u1ea5t (\u0111\u1ed1i di\u1ec7n v\u1edbi g\u00f3c t\u00f9) <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n l\u00e0: NP <\/span> <br\/> <span class='basic_green'> <i> Nh\u1eadn x\u00e9t: Qua b\u00e0i n\u00e0y \u0111\u1ec3 t\u00ecm c\u1ea1nh l\u1edbn nh\u1ea5t trong m\u1ed9t tam gi\u00e1c ta c\u00f3 th\u1ec3 t\u00ecm g\u00f3c l\u1edbn nh\u1ea5t trong tam gi\u00e1c \u0111\u00f3 sau \u0111\u00f3 \u00e1p \u0111\u1ee5ng \u0111\u1ecbnh l\u00fd quan h\u1ec7 gi\u1eefa g\u00f3c v\u00e0 c\u1ea1nh l\u1edbn nh\u1ea5t trong m\u1ed9t tam gi\u00e1c. <\/i> <\/span> "}]}],"id_ques":1824},{"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\u00f3 $\\widehat{A} \\geq 90^{o}$. Tr\u00ean c\u1ea1nh $AB$ v\u00e0 $AC$ l\u1ea7n l\u01b0\u1ee3t l\u1ea5y \u0111i\u1ec3m $E$ v\u00e0 $F$ (kh\u00f4ng tr\u00f9ng v\u1edbi c\u00e1c \u0111\u1ec9nh c\u1ee7a tam gi\u00e1c) . <br\/> Khi \u0111\u00f3 $BC > EF$. <b> \u0110\u00fang <\/b> hay <b> sai <\/b>? ","select":["A. \u0110\u00daNG ","B. SAI"],"hint":"So s\u00e1nh $EF$ v\u1edbi $EC$, so s\u00e1nh $EC$ v\u1edbi $BC$ r\u1ed3i so s\u00e1nh $EF$ v\u1edbi $BC$","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 $EF$ v\u00e0 $EC$ b\u1eb1ng c\u00e1ch x\u00e9t s\u1ed1 \u0111o g\u00f3c $\\widehat{EFC}$ (so s\u00e1nh v\u1edbi $90^{o}$) <br\/> <b> B\u01b0\u1edbc 2: <\/b> So s\u00e1nh $EC$ v\u00e0 $BC$ b\u1eb1ng c\u00e1ch x\u00e9t s\u1ed1 \u0111o g\u00f3c $\\widehat{BEC}$ (so s\u00e1nh v\u1edbi $90^{o}$) <br\/> <b> B\u01b0\u1edbc 3: <\/b> So s\u00e1nh $EF$ v\u1edbi $BC$ <br\/><br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai17/lv3/img\/H7C3B17_TB14.png' \/> <br\/> X\u00e9t $\\triangle{AEF}$ c\u00f3 $\\widehat{EFC}$ l\u00e0 g\u00f3c ngo\u00e0i t\u1ea1i \u0111\u1ec9nh $F$ <br\/> N\u00ean: $\\widehat{EFC} = \\widehat{A} + \\widehat{E_{1}}$ <br\/> M\u00e0 $\\widehat{A} \\geq 90^{o}$, suy ra $\\widehat{EFC}$ l\u00e0 g\u00f3c t\u00f9 <br\/> $\\triangle{EFC}$ c\u00f3 $\\widehat{EFC}$ l\u00e0 g\u00f3c t\u00f9 n\u00ean c\u1ea1nh $CE$ l\u00e0 c\u1ea1nh l\u1edbn nh\u1ea5t <br\/> $\\Rightarrow$ $EC > EF$ (1) <br\/> X\u00e9t $\\triangle{AEC}$ c\u00f3 $\\widehat{BEC}$ l\u00e0 g\u00f3c ngo\u00e0i t\u1ea1i \u0111\u1ec9nh $E$ <br\/> $\\Rightarrow$ $\\widehat{BEC} = \\widehat{A} + \\widehat{C_{1}}$ <br\/> M\u00e0 $\\widehat{A} \\geq 90^{o}$ n\u00ean $\\widehat{BEC}$ l\u00e0 g\u00f3c t\u00f9 <br\/> $\\triangle{BEC}$ c\u00f3 $\\widehat{BEC}$ l\u00e0 g\u00f3c t\u00f9 n\u00ean c\u1ea1nh $BC$ l\u00e0 c\u1ea1nh l\u1edbn nh\u1ea5t <br\/> $\\Rightarrow$ $BC > EC$ (2) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow$ $BC > EF$ <br\/> V\u1eady kh\u1eb3ng \u0111\u1ecbnh $BC > EF$ l\u00e0 <span class='basic_pink'>\u0110\u00daNG <\/span> ","column":2}]}],"id_ques":1825},{"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\u00f3 $\\widehat{A} = 3\\widehat{B} = 6\\widehat{C}$. K\u1ebb $AD$ vu\u00f4ng g\u00f3c v\u1edbi $BC$, $D$ thu\u1ed9c $BC$. <br\/> Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y \u0111\u00fang? ","select":["A. $AD < BD < CD $ ","B. $AD < CD < BD $","C. $BD < CD < AD$","D. $CD < AD < BD$ "],"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 c\u00e1c g\u00f3c c\u1ee7a $\\triangle{ABC}$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> T\u00ednh s\u1ed1 \u0111o $\\widehat{A_{1}}$, sau \u0111\u00f3 so s\u00e1nh $BD$ v\u1edbi $AD$ d\u1ef1a v\u00e0o m\u1ed1i quan h\u1ec7 gi\u1eefa g\u00f3c v\u00e0 c\u1ea1nh \u0111\u1ed1i di\u1ec7n trong $\\triangle{ABD}$ <br\/> <b> B\u01b0\u1edbc 3: <\/b> So s\u00e1nh $BD$ v\u00e0 $CD$ nh\u1edd b\u01b0\u1edbc trung gian l\u00e0 so s\u00e1nh $AB$ v\u1edbi $AC$ <br\/><br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai17/lv3/img\/H7C3B17_TB15.png' \/> <br\/> T\u1eeb $\\widehat{A} = 3\\widehat{B} = 6\\widehat{C}$ $\\Rightarrow$ $\\widehat{\\dfrac{A}{6}} = \\widehat{\\dfrac{B}{2}} = \\widehat{\\dfrac{C}{1}}$ <br\/> \u00c1p d\u1ee5ng t\u00ednh ch\u1ea5t d\u00e3y t\u1ec9 s\u1ed1 b\u1eb1ng nhau, ta c\u00f3: <br\/> $\\dfrac{\\widehat{A}}{6} = \\dfrac{\\widehat{B}}{2} = \\dfrac{\\widehat{C}}{1} = \\dfrac{\\widehat{A} + \\widehat{B} + \\widehat{C}}{6 + 2 + 1} = \\dfrac{180^{o}}{9} = 20^{o}$ <br\/> $\\Rightarrow$ $\\widehat{A} = 120^{o}; \\widehat{B} = 40^{o}; \\widehat{C} = 20^{o}$ <br\/> X\u00e9t $\\triangle{ACD}$ c\u00f3: <br\/> $\\begin{cases} \\widehat{ADC} = 90^{o} (gt) \\\\ \\widehat{C} = 20^{o} \\end{cases}$ <br\/> $\\Rightarrow$ $\\widehat{A_{2}} = 180^{o} - 90^{o} - 20^{o} = 70^{o}$ (t\u1ed5ng ba g\u00f3c trong m\u1ed9t tam gi\u00e1c) <br\/> $\\Rightarrow$ $\\widehat{A_{1}} = 120^{o} - 70^{o} = 50^{o}$ <br\/> X\u00e9t $\\triangle{ABD}$ c\u00f3: $\\widehat{A_{1}} > \\widehat{B}$ (v\u00ec $50^{o} > 40^{o}$) <br\/> $\\Rightarrow$ $BD > AD$ (1) <br\/> X\u00e9t $\\triangle{ABC}$ c\u00f3 $\\widehat{B} > \\widehat{C}$ (v\u00ec $40^{o} > 20^{o}$) <br\/> $\\Rightarrow$ $AB < AC$ $\\Rightarrow$ $AB^{2} < AC^{2}$ (2) <br\/> \u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pitago cho hai tam gi\u00e1c vu\u00f4ng $ADB$ v\u00e0 $ADC$ ta c\u00f3: <br\/> $AB^{2} = AD^{2} + DB^{2}$ (3) <br\/> $AC^{2} = AD^{2} + DC^{2}$ (4) <br\/> T\u1eeb (2), (3) v\u00e0 (4) $\\Rightarrow$ $AD^{2} + DB^{2} < AD^{2} + DC^{2}$ <br\/> $\\Leftrightarrow DB^{2} < DC^{2}$ <br\/> $\\Leftrightarrow DB < DC$ (5) <br\/> T\u1eeb (1) v\u00e0 (5) $\\Rightarrow$ $AD < BD < CD$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: A <\/span> ","column":2}]}],"id_ques":1826},{"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 $A$. G\u1ecdi $D$ l\u00e0 \u0111i\u1ec3m b\u1ea5t k\u00ec thu\u1ed9c mi\u1ec1n trong c\u1ee7a tam gi\u00e1c sao cho $\\widehat{ADB} > \\widehat{ADC}$. H\u00e3y so s\u00e1nh $DC$ v\u00e0 $DB$. ","select":["A. $DC > DB $ ","B. $DC = DB $","C. $DC < DB$"],"hint":"Tr\u00ean n\u1eeda m\u1eb7t ph\u1eb3ng kh\u00f4ng ch\u1ee9a \u0111i\u1ec3m $B$, b\u1edd l\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng $AC$ v\u1ebd th\u00eam tia $Ax$ sao cho $\\widehat{CAx} = \\widehat{BAD}$ <br\/> Tr\u00ean $Ax$ l\u1ea5y \u0111i\u1ec3m $E$ sao cho $AE = AD$ ","explain":" <span class='basic_left'> <span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> V\u1ebd th\u00eam \u0111\u01b0\u1eddng ph\u1ee5 (nh\u01b0 g\u1ee3i \u00fd) <br\/> <b> B\u01b0\u1edbc 2: <\/b> So s\u00e1nh $DB$ v\u00e0 $EC$ <br\/> <b> B\u01b0\u1edbc 3: <\/b> So s\u00e1nh $DC$ v\u00e0 $EC$ d\u1ef1a v\u00e0o quan h\u1ec7 gi\u1eefa g\u00f3c v\u00e0 c\u1ea1nh \u0111\u1ed1i di\u1ec7n trong $\\triangle{DEC}$ <br\/> \u0110\u1ec3 so s\u00e1nh \u0111\u01b0\u1ee3c $\\widehat{D_{2}}$ v\u00e0 $\\widehat{E_{2}}$ c\u1ea7n so s\u00e1nh $\\widehat{ADC}$ v\u00e0 $\\widehat{AEC}$ <br\/><br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai17/lv3/img\/H7C3B17_TB16.png' \/> <br\/> Tr\u00ean n\u1eeda m\u1eb7t ph\u1eb3ng kh\u00f4ng ch\u1ee9a \u0111i\u1ec3m $B$, b\u1edd l\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng $AC$, v\u1ebd tia $Ax$ sao cho $\\widehat{CAx} = \\widehat{BAD}$ <br\/> Tr\u00ean tia $Ax$ l\u1ea5y \u0111i\u1ec3m $E$ sao cho $AE = AD$ <br\/> X\u00e9t $\\triangle{ABD}$ v\u00e0 $\\triangle{ACE}$ c\u00f3: <br\/> $\\begin{cases} AD = AE (\\text{c\u00e1ch} \\hspace{0,2cm} \\text{l\u1ea5y} \\hspace{0,2cm} \\text{\u0111i\u1ec3m} \\hspace{0,2cm} E) \\\\ \\widehat{A_{1}} = \\widehat{A_{3}} (\\text{c\u00e1ch} \\hspace{0,2cm} \\text{v\u1ebd} \\hspace{0,2cm} Ax) \\\\ AB = AC (\\triangle{ABC} \\hspace{0,2cm} \\text{c\u00e2n}) \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{ADB} = \\triangle{AEC}$ (c. g. c) <br\/> $\\Rightarrow$ $BD = EC$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) (1) <br\/> $\\widehat{ADB} = \\widehat{AEC}$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) (2) <br\/> V\u00ec $\\widehat{ADB} > \\widehat{ADC}$ (gt) $\\Rightarrow$ $\\widehat{AEC} > \\widehat{ADC}$ <br\/> $\\Leftrightarrow$ $\\widehat{E_{1}} + \\widehat{E_{2}} > \\widehat{D_{1}} + \\widehat{D_{2}}$ (3) <br\/> M\u1eb7t kh\u00e1c ta c\u00f3: $\\widehat{D_{1}} = \\widehat{E_{1}}$ (do $AD = AE$ n\u00ean $\\triangle{ADE}$ c\u00e2n t\u1ea1i $A$) (4) <br\/> T\u1eeb (3) v\u00e0 (4) $\\Rightarrow$ $\\widehat{E_{2}} > \\widehat{D_{2}}$ <br\/> X\u00e9t $\\triangle{CDE}$ c\u00f3 $\\widehat{E_{2}} > \\widehat{D_{2}}$ (ch\u1ee9ng minh tr\u00ean) <br\/> N\u00ean $DC > EC$ (quan h\u1ec7 gi\u1eefa g\u00f3c v\u00e0 c\u1ea1nh \u0111\u1ed1i di\u1ec7n trong tam gi\u00e1c) (5) <br\/> T\u1eeb (1) v\u00e0 (5) $\\Rightarrow$ $DC > DB$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: A <\/span> ","column":3}]}],"id_ques":1827},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"Cho tam gi\u00e1c $ABC$ v\u00e0 tam gi\u00e1c $A'B'C'$ c\u00f3 $AB = A'B'$, $AC = A'C'$, $\\widehat{A} < \\widehat{A'}$ v\u00e0 $AB \\leq AC$ . H\u00e3y so s\u00e1nh $BC$ v\u00e0 $B'C'$. ","select":["A. $BC > B'C' $ ","B. $BC = B'C' $","C. $BC < B'C'$"],"hint":"","explain":" <span class='basic_left'> <span class='basic_green'>Ph\u00e2n t\u00edch b\u00e0i to\u00e1n:<\/span><br\/> \u0110\u1ec3 so s\u00e1nh $BC$ v\u00e0 $B'C'$ c\u1ea7n \u0111\u01b0a v\u1ec1 c\u00f9ng m\u1ed9t tam gi\u00e1c \u0111\u1ec3 \u00e1p d\u1ee5ng \u0111\u1ecbnh l\u00fd quan h\u1ec7 gi\u1eefa g\u00f3c v\u00e0 c\u1ea1nh \u0111\u1ed1i di\u1ec7n trong m\u1ed9t tam gi\u00e1c. \u0110\u1ec3 l\u00e0m \u0111\u01b0\u1ee3c vi\u1ec7c \u0111\u00f3 c\u1ea7n v\u1ebd th\u00eam \u0111\u01b0\u1eddng ph\u1ee5. <br\/> V\u1ebd $\\widehat{BAD} = \\widehat{A'}$ ($D$ n\u1eb1m tr\u00ean n\u1eeda m\u1eb7t ph\u1eb3ng b\u1edd l\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng $AC$ kh\u00f4ng ch\u1ee9a \u0111i\u1ec3m $B$). Khi \u0111\u00f3 ch\u1ee9ng minh \u0111\u01b0\u1ee3c $\\triangle{ABD} = \\triangle{A'B'C'}$. <br\/> B\u00e0i to\u00e1n chuy\u1ec3n v\u1ec1 so s\u00e1nh $BD$ v\u00e0 $BC$ trong $\\triangle{BCD}$ <br\/><br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai17/lv3/img\/H7C3B17_TB17.png' \/> <br\/> $\\blacktriangleright$ Tr\u00ean n\u1eeda m\u1eb7t ph\u1eb3ng b\u1edd $AC$ kh\u00f4ng ch\u1ee9a \u0111i\u1ec3m $B$ v\u1ebd tia $Ax$ sao cho $\\widehat{CAx} = \\widehat{A'} - \\widehat{A}$ <br\/> Khi \u0111\u00f3: $\\widehat{BAx} = \\widehat{BAC} + \\widehat{CAx} = \\widehat{A} + (\\widehat{A'} - \\widehat{A}) = \\widehat{A'}$ <br\/> V\u1ebd tia $Cy$ l\u00e0 tia \u0111\u1ed1i c\u1ee7a tia $CB$. Tr\u00ean tia $Ax$ l\u1ea5y \u0111i\u1ec3m $D$ sao cho: $AD = AC$ <br\/> $\\blacktriangleright$ $\\triangle{ABC}$ c\u00f3 $AB \\leq AC$ $\\Rightarrow$ $\\widehat{C} \\leq \\widehat{B}$ (quan h\u1ec7 gi\u1eefa g\u00f3c v\u00e0 c\u1ea1nh \u0111\u1ed1i di\u1ec7n trong m\u1ed9t tam gi\u00e1c) <br\/> $\\Rightarrow$ $\\widehat{C} \\leq 90^{o}$ (v\u00ec trong m\u1ed9t tam gi\u00e1c kh\u00f4ng c\u00f3 qu\u00e1 m\u1ed9t g\u00f3c t\u00f9) <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{ABD}$ v\u00e0 $\\triangle{A'B'C'}$ c\u00f3: <br\/> $\\begin{cases} AB = A'B' (gt) \\\\ \\widehat{BAD} = \\widehat{A'} (\\text{c\u00e1ch} \\hspace{0,2cm} \\text{v\u1ebd} \\hspace{0,2cm} Ax) \\\\ AD = A'C' (= AC) \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{ABD} = \\triangle{A'B'C'}$ (c. g. c) <br\/> $\\Rightarrow BD = B'C'$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) (1) <br\/> V\u00ec $AD = AC$ n\u00ean $\\triangle{ACD}$ c\u00e2n t\u1ea1i $A$ $\\Rightarrow$ $\\widehat{ACD} < 90^{o}$ <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{ABC}$ c\u00f3 $\\widehat{C} \\leq 90^{o}$ $\\Rightarrow$ $\\widehat{ACy} \\geq 90^{o}$ <br\/> $\\Rightarrow$ $\\widehat{ACD} < 90^{o} \\leq \\widehat{ACy}$ <br\/> Do \u0111\u00f3 tia $CD$ n\u1eb1m gi\u1eefa hai tia $CA$ v\u00e0 $Cy$ <br\/> $\\Rightarrow$ $A$ v\u00e0 $C$ kh\u00e1c ph\u00eda \u0111\u1ed1i v\u1edbi $DB$ <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{BCD}$ c\u00f3: <br\/> $\\widehat{BDC} = \\widehat{ADC} - \\widehat{ADB} < \\widehat{ADC} = \\widehat{ACD} < \\widehat{ACD} + \\widehat{ACB} = \\widehat{BCD}$ <br\/> $\\Rightarrow$ $BC < BD$ (quan h\u1ec7 gi\u1eefa g\u00f3c v\u00e0 c\u1ea1nh \u0111\u1ed1i di\u1ec7n trong m\u1ed9t tam gi\u00e1c) (2) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow$ $BC < B'C'$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: C <\/span> <br\/> <span class='basic_green'> <b> \u0110\u1ecbnh l\u00fd b\u1ed5 sung 1: <\/b> <i> N\u1ebfu hai tam gi\u00e1c c\u00f3 hai c\u1eb7p c\u1ea1nh b\u1eb1ng nhau nh\u01b0ng c\u1eb7p g\u00f3c xen gi\u1eefa kh\u00f4ng b\u1eb1ng nhau th\u00ec c\u1eb7p c\u1ea1nh \u0111\u1ed1i di\u1ec7n v\u1edbi n\u00f3 c\u0169ng kh\u00f4ng b\u1eb1ng nhau. C\u1ee5 th\u1ec3: C\u1ea1nh \u0111\u1ed1i di\u1ec7n v\u1edbi g\u00f3c l\u1edbn h\u01a1n l\u00e0 c\u1ea1nh l\u1edbn h\u01a1n <\/i> <\/span> ","column":3}]}],"id_ques":1828},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"Cho tam gi\u00e1c $ABC$ c\u00e2n t\u1ea1i \u0111\u1ec9nh $A$. Tr\u00ean $BC$ l\u1ea5y \u0111i\u1ec3m $M$ v\u00e0 $N$ sao cho $BM = MN = NC$. H\u00e3y so s\u00e1nh $\\widehat{BAM}$ v\u00e0 $\\widehat{MAN}$ ","select":["A. $\\widehat{BAM} > \\widehat{MAN} $ ","B. $\\widehat{BAM} =\\widehat{MAN} $","C. $\\widehat{BAM} < \\widehat{MAN} $"],"hint":"Tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia $MA$ v\u1ebd th\u00eam \u0111i\u1ec3m $D$ sao cho $MD = MA$","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 $AB$ v\u00e0 $AN$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> So s\u00e1nh $\\widehat{BAM}$ v\u00e0 $\\widehat{MAN}$ <br\/> \u0110\u1ec3 so s\u00e1nh \u0111\u01b0\u1ee3c ta c\u1ea7n \u0111\u01b0a v\u1ec1 c\u00f9ng m\u1ed9t tam gi\u00e1c. Do v\u1eady c\u1ea7n v\u1ebd th\u00eam \u0111i\u1ec3m $D$ tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia $MA$ sao cho $MD = MA$. <br\/> Khi \u0111\u00f3 b\u00e0i to\u00e1n chuy\u1ec3n v\u1ec1 so s\u00e1nh $\\widehat{ADN}$ v\u00e0 $\\widehat{NAD}$ trong $\\triangle{AND}$ <br\/><br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai17/lv3/img\/H7C3B17_TB18.png' \/> <br\/> $\\blacktriangleright$ $\\triangle{ANC}$ c\u00f3 $\\widehat{ANB}$ l\u00e0 g\u00f3c ngo\u00e0i t\u1ea1i \u0111\u1ec9nh $N$ <br\/> $\\Rightarrow$ $\\widehat{ANB} > \\widehat{C}$ (t\u00ednh ch\u1ea5t g\u00f3c ngo\u00e0i c\u1ee7a tam gi\u00e1c) <br\/> $\\Rightarrow$ $\\widehat{ANB} > \\widehat{B}$ (V\u00ec $\\widehat{B} = \\widehat{C}$ do $\\triangle{ABC}$ c\u00e2n t\u1ea1i $A$) <br\/> $\\blacktriangleright$ $\\triangle{ABN}$ c\u00f3 $\\widehat{ANB} > \\widehat{B}$ <br\/> $\\Rightarrow$ $AB > AN$ (quan h\u1ec7 gi\u1eefa g\u00f3c v\u00e0 c\u1ea1nh \u0111\u1ed1i di\u1ec7n trong m\u1ed9t tam gi\u00e1c) (1) <br\/> $\\blacktriangleright$ Tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia $MA$ l\u1ea5y \u0111i\u1ec3m $D$ sao cho: $MD = MA$ <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{MAB}$ v\u00e0 $\\triangle{MDN}$ c\u00f3: <br\/> $\\begin{cases} BM = MN (gt) \\\\ \\widehat{AMB} = \\widehat{DMN} (\\text{\u0111\u1ed1i} \\hspace{0,2cm} \\text{\u0111\u1ec9nh}) \\\\ AM = MD (\\text{c\u00e1ch} \\hspace{0,2cm} \\text{l\u1ea5y} \\hspace{0,2cm} \\text{\u0111i\u1ec3m} \\hspace{0,2cm} D) \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{MAB} = \\triangle{MDN}$ (c. g. c) <br\/> $\\Rightarrow$ $AB = DN$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) (2) <br\/> $\\widehat{MAB} = \\widehat{MDN}$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) (3) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow$ $AN < ND$ <br\/> $\\blacktriangleright$ $\\triangle{AND}$ c\u00f3 $AN < ND$ (ch\u1ee9ng minh tr\u00ean) n\u00ean $\\widehat{ADN} < \\widehat{NAD}$ (quan h\u1ec7 gi\u1eefa g\u00f3c v\u00e0 c\u1ea1nh \u0111\u1ed1i di\u1ec7n trong m\u1ed9t tam gi\u00e1c) <br\/> Hay $\\widehat{MDN} < \\widehat{MAN}$ (4) <br\/> T\u1eeb (3) v\u00e0 (4) $\\Rightarrow$ $\\widehat{BAM} < \\widehat{MAN}$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: C <\/span> <br\/> <span class='basic_green'> <b> <i> Nh\u1eadn x\u00e9t: Khi so s\u00e1nh c\u00e1c g\u00f3c n\u00ean chuy\u1ec3n v\u1ec1 so s\u00e1nh c\u00e1c g\u00f3c trong c\u00f9ng m\u1ed9t tam gi\u00e1c <br\/> Sai l\u1ea7m c\u00f3 th\u1ec3 m\u1eafc l\u00e0: L\u1ea7m t\u01b0\u1edfng $BM = MN$ th\u00ec $\\widehat{BAM} = \\widehat{MAN}$ <\/i> <\/span> ","column":3}]}],"id_ques":1829},{"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\u00f3 $AC > AB$. Tia ph\u00e2n gi\u00e1c g\u00f3c $A$ c\u1eaft $BC$ \u1edf $D$. G\u1ecdi $I$ l\u00e0 m\u1ed9t \u0111i\u1ec3m n\u1eb1m gi\u1eefa $A$ v\u00e0 $D$. Tr\u00ean c\u1ea1nh $AC$ l\u1ea5y \u0111i\u1ec3m $E$ sao cho $AE = AB$. <br\/> Khi \u0111\u00f3: $IC > IB$ <b> \u0111\u00fang <\/b> hay <b> sai <\/b>? ","select":["A. \u0110\u00daNG ","B. SAI"],"hint":"Chuy\u1ec3n b\u00e0i to\u00e1n v\u1ec1 so s\u00e1nh $IE$ v\u00e0 $IC$ trong $\\triangle{IEC}$","explain":" <span class='basic_left'> <img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai17/lv3/img\/H7C3B17_TB19.png' \/> <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{ABI}$ v\u00e0 $\\triangle{AEI}$ c\u00f3: <br\/> $\\begin{cases} AE = AB (gt) \\\\ \\widehat{A_{1}} = \\widehat{A_{2}} (gt) \\\\ \\text{c\u1ea1nh} \\hspace{0,2cm} AI \\hspace{0,2cm} \\text{chung} \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{ABI} = \\triangle{AEI}$ (c.g.c) <br\/> $\\Rightarrow$ $\\widehat{I_{1}} = \\widehat{I_{2}}$(hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) (1) <br\/> $\\blacktriangleright$ $\\widehat{IEC} > \\widehat{I_{1}}$ (g\u00f3c ngo\u00e0i t\u1ea1i $E$ c\u1ee7a $\\triangle{AIE}$) (2) <br\/> $\\blacktriangleright$ $\\widehat{I_{2}} > \\widehat{D_{1}}$ (g\u00f3c ngo\u00e0i t\u1ea1i $I$ c\u1ee7a $\\triangle{BID}$) (3) <br\/> $\\blacktriangleright$ $\\widehat{D_{1}} > \\widehat{ACD}$ (g\u00f3c ngo\u00e0i t\u1ea1i $D$ c\u1ee7a $\\triangle{ADC}$) (4) <br\/> $\\blacktriangleright$ $\\widehat{ACD} > \\widehat{C_{1}}$ (V\u00ec $I$ n\u1eb1m trong $\\triangle{ABC}$) (5) <br\/> T\u1eeb (1), (2), (3), (4), (5) $\\Rightarrow$ $\\widehat{IEC} > \\widehat{C_{1}}$ <br\/> $\\blacktriangleright$ $\\triangle{IEC}$ c\u00f3 $\\widehat{IEC} > \\widehat{C_{1}}$ $\\Rightarrow$ $IC > IE$ (6) <br\/> M\u1eb7t kh\u00e1c ta c\u00f3: $IB = IE$ (do $\\triangle{ABI} = \\triangle{AEI}$) (7) <br\/> T\u1eeb (6) v\u00e0 (7) $\\Rightarrow$ $IC > IB$ <br\/> V\u1eady kh\u1eb3ng \u0111\u1ecbnh $IC > IB$ l\u00e0 <span class='basic_pink'> \u0110\u00daNG <\/span> ","column":3}]}],"id_ques":1830}],"lesson":{"save":0,"level":3}}