{"segment":[{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"<span class='basic_left'> Cho h\u00ecnh v\u1ebd sau,<br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai16/lv1/img\/h934_D1.png' \/><\/center> <br\/> M\u1ec7nh \u0111\u1ec1 n\u00e0o sau \u0111\u00e2y l\u00e0 \u0111\u00fang?","select":["A. $\\widehat{AIC}=\\dfrac{\\text{s\u0111}\\overset\\frown{AD}+\\text{s\u0111}\\overset\\frown{BC}}{2}$ ","B. $\\widehat{AIC}=\\dfrac{\\text{s\u0111}\\overset\\frown{AC}+\\text{s\u0111}\\overset\\frown{BD}}{2}$ ","C. $\\widehat{AIC}=\\dfrac{\\text{s\u0111}\\overset\\frown{AD}-\\text{s\u0111}\\overset\\frown{BC}}{2}$ ","D. $\\widehat{AIC}=\\dfrac{\\text{s\u0111}\\overset\\frown{AC}-\\text{s\u0111}\\overset\\frown{BD}}{2}$"],"explain":" S\u1ed1 \u0111o c\u1ee7a g\u00f3c c\u00f3 \u0111\u1ec9nh \u1edf b\u00ean trong \u0111\u01b0\u1eddng tr\u00f2n b\u1eb1ng n\u1eeda t\u1ed5ng s\u1ed1 \u0111o hai cung b\u1ecb ch\u1eafn <br\/> $\\Rightarrow \\widehat{AIC}=\\dfrac{\\text{s\u0111}\\overset\\frown{AC}+\\text{s\u0111}\\overset\\frown{BD}}{2}$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B <\/span><\/span>","column":2}]}],"id_ques":1501},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","temp":"fill_the_blank","correct":[[["50"]]],"list":[{"point":5,"width":50,"type_input":"","ques":" <span class='basic_left'> Cho h\u00ecnh v\u1ebd: <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai16/lv1/img\/h934_D2.png' \/><\/center> Bi\u1ebft $\\widehat{BAC}={{30}^{o}};\\,\\widehat{BDC}={{55}^{o}}.$ S\u1ed1 \u0111o $\\overset\\frown{DmE}$ l\u00e0 _input_ $^o$","explain":" <span class='basic_left'>Ta c\u00f3: $\\widehat{BDC}=\\dfrac{1}{2}\\text{s\u0111}\\overset\\frown{BC}$ (\u0111\u1ecbnh l\u00ed g\u00f3c n\u1ed9i ti\u1ebfp) <br\/> $\\Rightarrow \\text{s\u0111}\\overset\\frown{BC}=2\\widehat{BDC}={{2.55}^{o}}={{110}^{o}}$ <br\/> $\\widehat{BAC}=\\dfrac{\\text{s\u0111}\\overset\\frown{BC}-\\text{s\u0111}\\overset\\frown{DmE}}{2}$ (\u0111\u1ecbnh l\u00ed g\u00f3c c\u00f3 \u0111\u1ec9nh \u1edf b\u00ean ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n) <br\/> $\\Rightarrow \\text{s\u0111}\\overset\\frown{DmE}=\\text{s\u0111}\\overset\\frown{BC}-2\\widehat{BAC}={{110}^{o}}-{{2.30}^{o}}={{50}^{o}}$ <br\/> <span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $50$ <\/span><\/span> "}]}],"id_ques":1502},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Cho h\u00ecnh v\u1ebd: <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai16/lv1/img\/h934_D3.png' \/><\/center><br\/> Bi\u1ebft $\\text{s\u0111}\\overset\\frown{BQ}={{42}^{o}};\\,\\text{s\u0111}\\overset\\frown{QD}={{38}^{o}}.$ T\u1ed5ng s\u1ed1 \u0111o hai g\u00f3c $BPD$ v\u00e0 $AQC$ l\u00e0:","select":["A. ${{80}^{o}}$ ","B. ${{62}^{o}}$ ","C. ${{40}^{o}}$ ","D. ${{36}^{o}}$"],"explain":"<span class='basic_left'> Ta c\u00f3: $\\text{s\u0111}\\overset\\frown{BD}=\\text{s\u0111}\\overset\\frown{BQ}+\\text{s\u0111}\\overset\\frown{QD}$ (do $Q$ n\u1eb1m gi\u1eefa $B$ v\u00e0 $D$) <br\/> $\\Rightarrow \\text{s\u0111}\\overset\\frown{BD}={{42}^{o}}+{{38}^{o}}={{80}^{o}}$ <br\/> $\\widehat{BPD}=\\dfrac{\\text{s\u0111}\\overset\\frown{BD}-\\text{s\u0111}\\overset\\frown{AC}}{2}$ (\u0111\u1ecbnh l\u00ed g\u00f3c c\u00f3 \u0111\u1ec9nh \u1edf ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n) <br\/> $\\widehat{AQC}=\\dfrac{1}{2}\\text{s\u0111}\\overset\\frown{AC}$ (\u0111\u1ecbnh l\u00ed g\u00f3c n\u1ed9i ti\u1ebfp) <br\/> $\\Rightarrow \\widehat{BPD}+\\widehat{AQC}=\\dfrac{1}{2}\\text{s\u0111}\\overset\\frown{AC}+\\dfrac{\\text{s\u0111}\\overset\\frown{BD}-\\text{s\u0111}\\overset\\frown{AC}}{2}\\\\=\\dfrac{\\text{s\u0111}\\overset\\frown{BD}}{2}\\\\=\\dfrac{{{80}^{o}}}{2}={{40}^{o}}$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C <\/span><\/span>","column":4}]}],"id_ques":1503},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Cho h\u00ecnh v\u1ebd, <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai16/lv1/img\/h934_D4.png' \/><\/center> <br\/> Bi\u1ebft tam gi\u00e1c $ABC$ c\u00e2n t\u1ea1i $A$, c\u00f3 $\\widehat{ACB}={{50}^{o}};\\,\\widehat{BCD}={{30}^{o}}.$ S\u1ed1 \u0111o $\\widehat{AQC}$ l\u00e0: ","select":["A. ${{160}^{o}}$ ","B. ${{80}^{o}}$ ","C. ${{40}^{o}}$ ","D. ${{20}^{o}}$ "],"explain":" <span class='basic_left'> Ta c\u00f3: $\\widehat{ABC}=\\widehat{ACB}={{50}^{o}}$ (t\u00ednh ch\u1ea5t tam gi\u00e1c c\u00e2n) <br\/> $\\left\\{ \\begin{align} & \\widehat{ABC}=\\dfrac{1}{2}\\text{s\u0111}\\overset\\frown{AC} \\\\ & \\widehat{BCD}=\\dfrac{1}{2}\\text{s\u0111}\\overset\\frown{BD} \\\\ \\end{align} \\right.$ (\u0111\u1ecbnh l\u00ed g\u00f3c n\u1ed9i ti\u1ebfp) <br\/> $ \\Rightarrow \\left\\{ \\begin{align} & \\text{s\u0111}\\overset\\frown{AC}=2.\\widehat{ABC}={{2.50}^{o}}={{100}^{o}} \\\\ & \\text{s\u0111}\\overset\\frown{BD}=2.\\widehat{BCD}={{2.30}^{o}}={{60}^{o}} \\\\ \\end{align} \\right. $ <br\/> $\\widehat{AQC}=\\dfrac{\\text{s\u0111}\\overset\\frown{BD}+\\text{s\u0111}\\overset\\frown{AC}}{2}$ (\u0111\u1ecbnh l\u00ed g\u00f3c c\u00f3 \u0111\u1ec9nh \u1edf trong \u0111\u01b0\u1eddng tr\u00f2n) <br\/> $\\Rightarrow \\widehat{AQC}=\\dfrac{{{100}^{o}}+{{60}^{o}}}{2}={{80}^{o}}$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B <\/span><\/span>","column":4}]}],"id_ques":1504},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","temp":"fill_the_blank","correct":[[["30"]]],"list":[{"point":5,"width":50,"type_input":"","ques":" <span class='basic_left'> Cho \u0111\u01b0\u1eddng tr\u00f2n $(O)$ \u0111\u01b0\u1eddng k\u00ednh $AB$, v\u1ebd c\u00e1t tuy\u1ebfn $BD$ sao cho $\\text{s\u0111}\\overset\\frown{BD}={{60}^{o}}$ v\u00e0 ti\u1ebfp tuy\u1ebfn t\u1ea1i $A$ c\u1eaft \u0111\u01b0\u1eddng th\u1eb3ng $BD$ t\u1ea1i $M$. T\u00ednh g\u00f3c $AMB$. <br\/> <b> \u0110\u00e1p s\u1ed1: <\/b> $\\widehat{AMB}=$_input_ $^o$","hint":"S\u1ed1 \u0111o c\u1ee7a g\u00f3c c\u00f3 \u0111\u1ec9nh n\u1eb1m ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n b\u1eb1ng n\u1eeda hi\u1ec7u s\u1ed1 \u0111o hai cung b\u1ecb ch\u1eafn","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai16/lv1/img\/h934_D5.png' \/><\/center> <br\/> Ta c\u00f3: $\\text{s\u0111}\\overset\\frown{AmB}=\\text{s\u0111}\\overset\\frown{ADB}={{180}^{o}}$ ($AB$ l\u00e0 \u0111\u01b0\u1eddng k\u00ednh) <br\/> $\\text{s\u0111}\\overset\\frown{AD}+\\text{s\u0111}\\overset\\frown{DB}=\\text{s\u0111}\\overset\\frown{ADB}$ ($D$ n\u1eb1m gi\u1eefa $A$ v\u00e0 $B$) <br\/> $\\Rightarrow \\text{s\u0111}\\overset\\frown{AD}={{180}^{o}}-\\text{s\u0111}\\overset\\frown{BD}={{180}^{o}}-{{60}^{o}}={{120}^{o}}$ <br\/> L\u1ea1i c\u00f3: $\\widehat{AMB}=\\dfrac{\\text{s\u0111}\\overset\\frown{AmB}-\\text{s\u0111}\\overset\\frown{AD}}{2}$ (\u0111\u1ecbnh l\u00ed g\u00f3c c\u00f3 \u0111\u1ec9nh n\u1eb1m ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n) <br\/> $\\Rightarrow \\widehat{AMB}=\\dfrac{{{180}^{o}}-{{120}^{o}}}{2}={{30}^{o}}$ <br\/> <span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $30$ <\/span><\/span> "}]}],"id_ques":1505},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","temp":"fill_the_blank","correct":[[["30"]]],"list":[{"point":5,"width":50,"type_input":"","ques":" <span class='basic_left'> Cho \u0111\u01b0\u1eddng tr\u00f2n $(O)$, \u0111\u01b0\u1eddng k\u00ednh $BC$. L\u1ea5y \u0111i\u1ec3m $A$ thu\u1ed9c \u0111\u01b0\u1eddng tr\u00f2n sao cho $\\text{s\u0111}\\overset\\frown{AB}={{120}^{o}}$. Ti\u1ebfp tuy\u1ebfn t\u1ea1i $A$ c\u1eaft $BC$ t\u1ea1i $M$. S\u1ed1 \u0111o g\u00f3c $M$ l\u00e0_input_ $^o$","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai16/lv1/img\/h934_D6.png' \/><\/center> <br\/> Ta c\u00f3 $\\text{s\u0111}\\overset\\frown{AC}+\\text{s\u0111}\\overset\\frown{AB}=\\text{s\u0111}\\overset\\frown{BC}={{180}^{o}}$ (do $A$ n\u1eb1m gi\u1eefa $B$ v\u00e0 $C$; $BC$ l\u00e0 \u0111\u01b0\u1eddng k\u00ednh) <br\/> $\\Rightarrow \\text{s\u0111}\\overset\\frown{AC}={{180}^{o}}-\\text{s\u0111}\\overset\\frown{AB}\\\\={{180}^{o}}-{{120}^{o}}={{60}^{o}}$ <br\/> $\\widehat{M}=\\dfrac{\\text{s\u0111}\\overset\\frown{AB}-\\text{s\u0111}\\overset\\frown{AC}}{2}$ (\u0111\u1ecbnh l\u00ed g\u00f3c c\u00f3 \u0111\u1ec9nh n\u1eb1m ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n) <br\/> $\\Rightarrow \\widehat{M}=\\dfrac{{{120}^{o}}-{{60}^{o}}}{2}={{30}^{o}}$<br\/> <span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $30$ <\/span><\/span> "}]}],"id_ques":1506},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","temp":"fill_the_blank","correct":[[["90"]]],"list":[{"point":5,"width":50,"type_input":"","ques":" <span class='basic_left'> Cho \u0111\u01b0\u1eddng tr\u00f2n $(O)$, hai d\u00e2y $AB$ v\u00e0 $CD$ c\u1eaft nhau t\u1ea1i $I$ sao cho $\\widehat{DIB}={{40}^{o}}$ ($I$ n\u1eb1m ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n). Hai \u0111o\u1ea1n th\u1eb3ng $BC$ v\u00e0 $AD$ c\u1eaft nhau t\u1ea1i $K$. Bi\u1ebft $\\text{s\u0111}\\overset\\frown{DB}={{130}^{o}},$ t\u00ednh g\u00f3c $BKD$. <br\/> <b> \u0110\u00e1p s\u1ed1: <\/b> $\\widehat{BKD} =$ _input_ $^o$ ","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai16/lv1/img\/h934_D7.png' \/><\/center> <br\/> Ta c\u00f3: $\\widehat{DIB}=\\dfrac{\\text{s\u0111}\\overset\\frown{BD}-\\text{s\u0111}\\overset\\frown{AC}}{2}$ (\u0111\u1ecbnh l\u00ed g\u00f3c c\u00f3 \u0111\u1ec9nh n\u1eb1m ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n) <br\/> $\\Rightarrow \\text{s\u0111}\\overset\\frown{AC}=\\text{s\u0111}\\overset\\frown{BD}-2\\widehat{DIB}\\\\ ={{130}^{o}}-{{2.40}^{o}}={{50}^{o}}$ <br\/> $\\widehat{DKB}=\\dfrac{\\text{s\u0111}\\overset\\frown{BD}+\\text{s\u0111}\\overset\\frown{AC}}{2}$ (\u0111\u1ecbnh l\u00ed g\u00f3c c\u00f3 \u0111\u1ec9nh n\u1eb1m trong \u0111\u01b0\u1eddng tr\u00f2n) <br\/> $\\Rightarrow \\widehat{DKB}=\\dfrac{{{130}^{o}}+{{50}^{o}}}{2}={{90}^{o}}$ <br\/> <span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $90$ <\/span><\/span> "}]}],"id_ques":1507},{"time":24,"part":[{"time":3,"title":"<span class='basic_left'> Cho \u0111\u01b0\u1eddng tr\u00f2n $(O)$ v\u00e0 d\u00e2y $AB$ l\u1edbn h\u01a1n d\u00e2y $AC$. G\u1ecdi $M, N$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa c\u1ee7a $\\overset\\frown{AB}$ v\u00e0 $\\overset\\frown{AC}.$ \u0110\u01b0\u1eddng th\u1eb3ng $MN$ l\u1ea7n l\u01b0\u1ee3t c\u1eaft d\u00e2y $AB$ v\u00e0 $AC$ t\u1ea1i $E$ v\u00e0 $H$. Ch\u1ee9ng minh $\\widehat{BEN}=\\widehat{MHC}$.","title_trans":"S\u1eafp x\u1ebfp c\u00e1c c\u00e2u \u0111\u1ec3 \u0111\u01b0\u1ee3c b\u00e0i ch\u1ee9ng minh","temp":"sequence","correct":[[[3],[1],[4],[2]]],"list":[{"point":5,"left":["$\\Rightarrow \\widehat{AEN}=\\widehat{AHM}$","Ta c\u00f3: $\\overset\\frown{AN}=\\overset\\frown{NC};\\,\\overset\\frown{AM}=\\overset\\frown{MB}$ (gi\u1ea3 thi\u1ebft)","$\\Rightarrow \\widehat{BEN}=\\widehat{MHC}$ (c\u00f9ng b\u00f9 v\u1edbi hai g\u00f3c b\u1eb1ng nhau) ","M\u00e0 $\\left\\{ \\begin{align} & \\widehat{AEN}=\\dfrac{\\text{s\u0111}\\overset\\frown{AN}+\\text{s\u0111}\\overset\\frown{MB}}{2} \\\\ & \\widehat{AHM}=\\dfrac{\\text{s\u0111}\\overset\\frown{AM}+\\text{s\u0111}\\overset\\frown{NC}}{2} \\\\ \\end{align} \\right.$ (\u0111\u1ecbnh l\u00ed g\u00f3c c\u00f3 \u0111\u1ec9nh n\u1eb1m trong \u0111\u01b0\u1eddng tr\u00f2n)"],"top":120,"explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai16/lv1/img\/h934_D8.png' \/><\/center> <br\/> Ta c\u00f3: $\\overset\\frown{AN}=\\overset\\frown{NC};\\,\\overset\\frown{AM}=\\overset\\frown{MB}$ (gi\u1ea3 thi\u1ebft) <br\/> M\u00e0 $\\left\\{ \\begin{align} & \\widehat{AEN}=\\dfrac{\\text{s\u0111}\\overset\\frown{AN}+\\text{s\u0111}\\overset\\frown{MB}}{2} \\\\ & \\widehat{AHM}=\\dfrac{\\text{s\u0111}\\overset\\frown{AM}+\\text{s\u0111}\\overset\\frown{NC}}{2} \\\\ \\end{align} \\right.$ (\u0111\u1ecbnh l\u00ed g\u00f3c c\u00f3 \u0111\u1ec9nh n\u1eb1m trong \u0111\u01b0\u1eddng tr\u00f2n) <br\/> $\\Rightarrow \\widehat{AEN}=\\widehat{AHM}$ <br\/> $\\Rightarrow \\widehat{BEN}=\\widehat{MHC}$ (c\u00f9ng b\u00f9 v\u1edbi hai g\u00f3c b\u1eb1ng nhau) <\/span>"}]}],"id_ques":1508},{"time":24,"part":[{"time":3,"title":"<span class='basic_left'> Cho \u0111\u01b0\u1eddng tr\u00f2n $(O)$ \u0111\u01b0\u1eddng k\u00ednh $AB$, ti\u1ebfp tuy\u1ebfn t\u1ea1i $A$ c\u1eaft c\u00e1t tuy\u1ebfn $BD$ t\u1ea1i $M$. Ti\u1ebfp tuy\u1ebfn t\u1ea1i $D$ c\u1eaft $AM$ t\u1ea1i $Q$. Ch\u1ee9ng minh tam gi\u00e1c $DQM$ c\u00e2n t\u1ea1i $Q$.","title_trans":"S\u1eafp x\u1ebfp c\u00e1c c\u00e2u \u0111\u1ec3 \u0111\u01b0\u1ee3c b\u00e0i ch\u1ee9ng minh","temp":"sequence","correct":[[[3],[5],[4],[2],[1]]],"list":[{"point":5,"left":["M\u00e0 $\\widehat{BDx}=\\widehat{MDQ}$ (hai g\u00f3c \u0111\u1ed1i \u0111\u1ec9nh) $\\Rightarrow \\widehat{DAB}=\\widehat{MDQ}$ "," $\\Rightarrow \\Delta MDQ$ c\u00e2n t\u1ea1i $Q$ (d\u1ea5u hi\u1ec7u nh\u1eadn bi\u1ebft)","M\u1eb7t kh\u00e1c $\\widehat{DAB}=\\widehat{QMD}$ (c\u00f9ng ph\u1ee5 v\u1edbi $\\widehat{MAD}$) $\\Rightarrow \\widehat{MDQ}=\\widehat{QMD}$","$\\Rightarrow \\widehat{MDA}=90^o$ (hai g\u00f3c k\u1ec1 b\u00f9)","Ta c\u00f3: $\\widehat{DAB}=\\widehat{BDx}$ (g\u00f3c n\u1ed9i ti\u1ebfp v\u00e0 g\u00f3c t\u1ea1o b\u1edfi tia ti\u1ebfp tuy\u1ebfn v\u00e0 d\u00e2y cung c\u00f9ng ch\u1eafn $\\overset\\frown{BD}$) <br\/> $\\widehat{ADB}=90^o$ (g\u00f3c n\u1ed9i ti\u1ebfp ch\u1eafn n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n)"],"top":100,"explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai16/lv1/img\/h934_D9.png' \/><\/center> <br\/> Ta c\u00f3: $\\widehat{DAB}=\\widehat{BDx}$ (g\u00f3c n\u1ed9i ti\u1ebfp v\u00e0 g\u00f3c t\u1ea1o b\u1edfi tia ti\u1ebfp tuy\u1ebfn v\u00e0 d\u00e2y cung c\u00f9ng ch\u1eafn $\\overset\\frown{BD}$) <br\/> $\\widehat{ADB}=90^o$ (g\u00f3c n\u1ed9i ti\u1ebfp ch\u1eafn n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n) <br\/> $\\Rightarrow \\widehat{MDA}=90^o$ (hai g\u00f3c k\u1ec1 b\u00f9) <br\/> M\u00e0 $\\widehat{BDx}=\\widehat{MDQ}$ (hai g\u00f3c \u0111\u1ed1i \u0111\u1ec9nh) <br\/> $\\Rightarrow \\widehat{DAB}=\\widehat{MDQ}$ <br\/> M\u1eb7t kh\u00e1c $\\widehat{DAB}=\\widehat{QMD}$ (c\u00f9ng ph\u1ee5 v\u1edbi $\\widehat{MAD}$) <br\/> $\\Rightarrow \\widehat{MDQ}=\\widehat{QMD}$ <br\/> $\\Rightarrow \\Delta MDQ$ c\u00e2n t\u1ea1i $Q$ (d\u1ea5u hi\u1ec7u nh\u1eadn bi\u1ebft) <\/span>"}]}],"id_ques":1509},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"<span class='basic_left'> Cho h\u00ecnh v\u1ebd, <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai16/lv1/img\/h934_D10.1.png' \/><\/center><br\/> Ba b\u1ea1n An, B\u00ecnh, C\u01b0\u1eddng \u0111\u1ee9ng \u1edf ba v\u1ecb tr\u00ed $A, \\,B, \\,C$ nh\u01b0 h\u00ecnh v\u1ebd \u0111\u1ec3 \u0111\u00e1 b\u00f3ng v\u00e0o c\u1ea7u m\u00f4n $DE$. So s\u00e1nh s\u1ed1 \u0111o c\u00e1c g\u00f3c s\u00fat $DAE,\\,DBE,\\,DCE.$ ","select":["A. $\\widehat{DAE}>\\widehat{DCE}>\\widehat{DBE}$ ","B. $\\widehat{DBE}>\\widehat{DCE}>\\widehat{DAE}$ ","C. $\\widehat{DAE}>\\widehat{DBE}>\\widehat{DCE}$ ","D. $\\widehat{DBE}>\\widehat{DAE}>\\widehat{DCE}$"],"explain":"<span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai16/lv1/img\/h934_D10.2.png' \/><\/center><br\/> Ta c\u00f3: $\\widehat{DAE}>\\dfrac{\\text{s\u0111}\\overset\\frown{DE}}{2}$ (do $A$ n\u1eb1m trong \u0111\u01b0\u1eddng tr\u00f2n) <br\/> $\\widehat{DBE}=\\dfrac{\\text{s\u0111}\\overset\\frown{DE}}{2}$ (\u0111\u1ecbnh l\u00ed g\u00f3c n\u1ed9i ti\u1ebfp) <br\/> $\\widehat{DCE}<\\dfrac{\\text{s\u0111}\\overset\\frown{DE}}{2}$ (do $C$ n\u1eb1m ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n) <br\/> $\\Rightarrow \\widehat{DAE}>\\widehat{DBE}>\\widehat{DCE}$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C <\/span><\/span>","column":2}]}],"id_ques":1510},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","temp":"fill_the_blank","correct":[[["50"],["130"]]],"list":[{"point":5,"width":70,"type_input":"","ques":" <span class='basic_left'> Cho n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n $(O)$ \u0111\u01b0\u1eddng k\u00ednh $AB$, cung $CD$ c\u00f3 s\u1ed1 \u0111o b\u1eb1ng $80^o$ ($D$ thu\u1ed9c cung $BC$). G\u1ecdi $E$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $AC$ v\u00e0 $BD$, $F$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $AD$ v\u00e0 $BC$. <br\/> Khi \u0111\u00f3: $\\widehat{AEB}=$ _input_ $^o$; $\\widehat{AFB}=$ _input_ $^o$","explain":" <span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai16/lv1/img\/h934_D11.png' \/><\/center> <br\/> Ta c\u00f3: $\\text{s\u0111}\\overset\\frown{AB}={{180}^{o}}$ ($AB$ l\u00e0 \u0111\u01b0\u1eddng k\u00ednh) <br\/> $\\widehat{AEB}=\\dfrac{\\text{s\u0111}\\overset\\frown{AB}-\\text{s\u0111}\\overset\\frown{CD}}{2}$ (\u0111\u1ecbnh l\u00ed g\u00f3c c\u00f3 \u0111\u1ec9nh n\u1eb1m ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n) <br\/> $\\Rightarrow \\widehat{AEB}=\\dfrac{{{180}^{o}}-{{80}^{o}}}{2}={{50}^{o}}$ <br\/>$\\widehat{AFB}=\\dfrac{\\text{s\u0111}\\overset\\frown{AB}+\\text{s\u0111}\\overset\\frown{CD}}{2}$ (\u0111\u1ecbnh l\u00ed g\u00f3c c\u00f3 \u0111\u1ec9nh n\u1eb1m trong \u0111\u01b0\u1eddng tr\u00f2n) <br\/> $\\Rightarrow \\widehat{AFB}=\\dfrac{{{180}^{o}}+{{80}^{o}}}{2}={{130}^{o}}$ <br\/> <span class='basic_pink'> V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u1ea7n l\u01b0\u1ee3t l\u00e0 $50$ v\u00e0 $130$ <\/span><\/span> "}]}],"id_ques":1511},{"time":24,"part":[{"time":3,"title":"<span class='basic_left'> Cho \u0111\u01b0\u1eddng tr\u00f2n $(O)$ \u0111\u01b0\u1eddng k\u00ednh $AB$, d\u00e2y $MN$ vu\u00f4ng g\u00f3c v\u1edbi $AB$. \u0110i\u1ec3m $C$ thu\u1ed9c cung $BM$. Ti\u1ebfp tuy\u1ebfn t\u1ea1i $C$ c\u1eaft \u0111\u01b0\u1eddng th\u1eb3ng $MN$ t\u1ea1i $K$, $AC$ c\u1eaft $MN$ t\u1ea1i $E$. Ch\u1ee9ng minh tam gi\u00e1c $KEC$ c\u00e2n.","title_trans":"S\u1eafp x\u1ebfp c\u00e1c c\u00e2u \u0111\u1ec3 \u0111\u01b0\u1ee3c b\u00e0i ch\u1ee9ng minh","temp":"sequence","correct":[[[3],[1],[5],[4],[2]]],"list":[{"point":5,"left":["Ta c\u00f3: $\\widehat{KCE}=\\dfrac{1}{2}\\text{s\u0111}\\overset\\frown{AC}$ (\u0111\u1ecbnh l\u00ed g\u00f3c t\u1ea1o b\u1edfi tia ti\u1ebfp tuy\u1ebfn v\u00e0 d\u00e2y cung) ","Do $AB\\bot MN$ <br\/> $\\Rightarrow AB$ l\u00e0 trung tr\u1ef1c c\u1ee7a $MN$ (\u0111\u1ecbnh l\u00ed \u0111\u01b0\u1eddng k\u00ednh v\u00e0 d\u00e2y cung) "," $\\Rightarrow \\Delta KEC$ c\u00e2n (d\u1ea5u hi\u1ec7u nh\u1eadn bi\u1ebft) "," $\\Rightarrow \\widehat{KCE}=\\dfrac{\\text{s\u0111}\\overset\\frown{AM}+\\text{s\u0111}\\overset\\frown{MC}}{2}=\\dfrac{\\text{s\u0111}\\overset\\frown{AN}+\\text{s\u0111}\\overset\\frown{MC}}{2}=\\widehat{CEK}$","$\\Rightarrow AM=AN$ <br\/> $\\Rightarrow \\overset\\frown{AM}=\\overset\\frown{AN}$ (\u0111\u1ecbnh l\u00ed li\u00ean h\u1ec7 gi\u1eefa cung v\u00e0 d\u00e2y)"],"top":75,"explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai16/lv1/img\/h934_D12.png' \/><\/center> <br\/> Do $AB\\bot MN$ <br\/> $\\Rightarrow AB$ l\u00e0 trung tr\u1ef1c c\u1ee7a $MN$ (\u0111\u1ecbnh l\u00ed \u0111\u01b0\u1eddng k\u00ednh v\u00e0 d\u00e2y cung) <br\/> $\\Rightarrow AM=AN$ <br\/> $\\Rightarrow \\overset\\frown{AM}=\\overset\\frown{AN}$ (\u0111\u1ecbnh l\u00ed li\u00ean h\u1ec7 gi\u1eefa cung v\u00e0 d\u00e2y) <br\/> Ta c\u00f3: $\\widehat{KCE}=\\dfrac{1}{2}\\text{s\u0111}\\overset\\frown{AC}$ (\u0111\u1ecbnh l\u00ed g\u00f3c t\u1ea1o b\u1edfi tia ti\u1ebfp tuy\u1ebfn v\u00e0 d\u00e2y cung) <br\/> $\\Rightarrow \\widehat{KCE}=\\dfrac{\\text{s\u0111}\\overset\\frown{AM}+\\text{s\u0111}\\overset\\frown{MC}}{2}=\\dfrac{\\text{s\u0111}\\overset\\frown{AN}+\\text{s\u0111}\\overset\\frown{MC}}{2}=\\widehat{CEK}$ <br\/> $\\Rightarrow \\Delta KEC$ c\u00e2n (d\u1ea5u hi\u1ec7u nh\u1eadn bi\u1ebft) <\/span>"}]}],"id_ques":1512},{"time":24,"part":[{"time":3,"title":"<span class='basic_left'> Cho \u0111\u01b0\u1eddng tr\u00f2n $(O)$. Qua \u0111i\u1ec3m $A$ n\u1eb1m ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n, k\u1ebb ti\u1ebfp tuy\u1ebfn $AB$ v\u00e0 c\u00e1t tuy\u1ebfn $ACD$. Qua $B$ k\u1ebb \u0111\u01b0\u1eddng vu\u00f4ng g\u00f3c v\u1edbi tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $A$, c\u1eaft $CD$ t\u1ea1i $E$ v\u00e0 c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n \u1edf $M$ (kh\u00e1c $B$). Ch\u1ee9ng minh $M$ l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa cung $CD$. ","title_trans":"S\u1eafp x\u1ebfp c\u00e1c c\u00e2u \u0111\u1ec3 \u0111\u01b0\u1ee3c b\u00e0i ch\u1ee9ng minh","temp":"sequence","correct":[[[2],[5],[4],[1],[3]]],"list":[{"point":5,"left":["$\\Rightarrow \\Delta ABE$ c\u00e2n t\u1ea1i $A$ (d\u1ea5u hi\u1ec7u nh\u1eadn bi\u1ebft) $\\Rightarrow \\widehat{ABE}=\\widehat{AEB}$ (t\u00ednh ch\u1ea5t tam gi\u00e1c c\u00e2n)(1) ","T\u1eeb (1), (2) v\u00e0 (3) suy ra $\\overset\\frown{CM}=\\overset\\frown{DM}$ $\\Rightarrow M$ l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa cung $CD$ "," $\\widehat{ABE}=\\dfrac{\\text{s\u0111}\\overset\\frown{BC}+\\text{s\u0111}\\overset\\frown{CM}}{2}=\\dfrac{\\text{s\u0111}\\overset\\frown{BCM}}{2}$ (\u0111\u1ecbnh l\u00ed g\u00f3c t\u1ea1o b\u1edfi tia ti\u1ebfp tuy\u1ebfn v\u00e0 d\u00e2y cung) (3) "," Ta c\u00f3: $AH$ v\u1eeba l\u00e0 \u0111\u01b0\u1eddng cao v\u1eeba l\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a tam gi\u00e1c $ABE$","M\u00e0 $\\widehat{AEB}=\\dfrac{\\text{s\u0111}\\overset\\frown{BC}+\\text{s\u0111}\\overset\\frown{DM}}{2}$ (\u0111\u1ecbnh l\u00ed g\u00f3c c\u00f3 \u0111\u1ec9nh n\u1eb1m trong \u0111\u01b0\u1eddng tr\u00f2n) (2)"],"top":75,"explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai16/lv1/img\/h934_D13.png' \/><\/center> <br\/> Ta c\u00f3: $AH$ v\u1eeba l\u00e0 \u0111\u01b0\u1eddng cao v\u1eeba l\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a tam gi\u00e1c $ABE$ <br\/> $\\Rightarrow \\Delta ABE$ c\u00e2n t\u1ea1i $A$ (d\u1ea5u hi\u1ec7u nh\u1eadn bi\u1ebft) <br\/> $\\Rightarrow \\widehat{ABE}=\\widehat{AEB}$ (t\u00ednh ch\u1ea5t tam gi\u00e1c c\u00e2n) (1) <br\/> M\u00e0 $\\widehat{AEB}=\\dfrac{\\text{s\u0111}\\overset\\frown{BC}+\\text{s\u0111}\\overset\\frown{DM}}{2}$ (\u0111\u1ecbnh l\u00ed g\u00f3c c\u00f3 \u0111\u1ec9nh n\u1eb1m trong \u0111\u01b0\u1eddng tr\u00f2n) (2) <br\/> $\\widehat{ABE}=\\dfrac{\\text{s\u0111}\\overset\\frown{BC}+\\text{s\u0111}\\overset\\frown{CM}}{2}=\\dfrac{\\text{s\u0111}\\overset\\frown{BCM}}{2}$ (\u0111\u1ecbnh l\u00ed g\u00f3c t\u1ea1o b\u1edfi tia ti\u1ebfp tuy\u1ebfn v\u00e0 d\u00e2y cung) (3) <br\/> T\u1eeb (1), (2) v\u00e0 (3) suy ra $\\overset\\frown{CM}=\\overset\\frown{DM}$ <br\/> $\\Rightarrow M$ l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa cung $CD$ <\/span>"}]}],"id_ques":1513},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"<span class='basic_left'> Cho \u0111\u01b0\u1eddng tr\u00f2n $(O)$, \u0111\u01b0\u1eddng k\u00ednh $AB$. L\u1ea5y \u0111i\u1ec3m $N$ thu\u1ed9c \u0111\u01b0\u1eddng tr\u00f2n sao cho $\\widehat{BAN}={{24}^{o}}$. Ti\u1ebfp tuy\u1ebfn t\u1ea1i $N$ c\u1eaft $AB$ t\u1ea1i $M$. S\u1ed1 \u0111o $\\widehat{AMN}$ l\u00e0: ","select":["A. ${{50}^{o}}$ ","B. ${{60}^{o}}$ ","C. ${{42}^{o}}$ ","D. ${{45}^{o}}$"],"explain":"<span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai16/lv1/img\/h934_D14.png' \/><\/center><br\/> Ta c\u00f3: $\\widehat{ONM}={{90}^{o}}$ (t\u00ednh ch\u1ea5t ti\u1ebfp tuy\u1ebfn) <br\/> $\\widehat{MON}=2\\widehat{NAB}$ (g\u00f3c n\u1ed9i ti\u1ebfp v\u00e0 g\u00f3c \u1edf t\u00e2m c\u00f9ng ch\u1eafn m\u1ed9t cung) <br\/> $\\Rightarrow \\widehat{MON}={{2.24}^{o}}={{48}^{o}}$ <br\/> M\u00e0 $\\widehat{MON}+\\widehat{NMO}={{90}^{o}}$ (t\u1ed5ng hai g\u00f3c nh\u1ecdn trong tam gi\u00e1c vu\u00f4ng $OMN$) <br\/> $\\Rightarrow \\widehat{NMO}=\\widehat{AMN}={{90}^{o}}-{{48}^{o}}={{42}^{o}}$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C <\/span><\/span>","column":4}]}],"id_ques":1514},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"<span class='basic_left'> Cho \u0111\u01b0\u1eddng tr\u00f2n $(O)$, b\u1ed1n \u0111i\u1ec3m $A, \\,B, \\,C, \\,D$ thu\u1ed9c \u0111\u01b0\u1eddng tr\u00f2n sao cho $\\widehat{BCA}={{45}^{o}};\\,\\widehat{CAD}={{30}^{o}}.$ \u0110o\u1ea1n th\u1eb3ng $AC$ c\u1eaft $BD$ t\u1ea1i $E$. S\u1ed1 \u0111o g\u00f3c $\\widehat{AEB}$ l\u00e0: ","select":["A. ${{70}^{o}}$ ","B. ${{75}^{o}}$ ","C. ${{90}^{o}}$ ","D. ${{60}^{o}}$"],"explain":"<span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai16/lv1/img\/h934_D15.png' \/><\/center><br\/> Ta c\u00f3: $\\text{s\u0111}\\overset\\frown{AB}=2\\widehat{ACB}$ (\u0111\u1ecbnh l\u00ed g\u00f3c n\u1ed9i ti\u1ebfp) <br\/> $\\Rightarrow \\text{s\u0111}\\overset\\frown{AB}={{2.45}^{o}}={{90}^{o}}$ <br\/> T\u01b0\u01a1ng t\u1ef1 $\\text{s\u0111}\\overset\\frown{CD}={{2.30}^{o}}={{60}^{o}}$ <br\/> M\u00e0 $\\widehat{AEB}=\\dfrac{\\text{s\u0111}\\overset\\frown{AB}+\\text{s\u0111}\\overset\\frown{CD}}{2}$ (\u0111\u1ecbnh l\u00ed g\u00f3c c\u00f3 \u0111\u1ec9nh n\u1eb1m trong \u0111\u01b0\u1eddng tr\u00f2n) <br\/> $\\Rightarrow \\widehat{AEB}=\\dfrac{{{90}^{o}}+{{60}^{o}}}{2}={{75}^{o}}$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B <\/span><\/span>","column":4}]}],"id_ques":1515},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"<span class='basic_left'> Cho tam gi\u00e1c $ABC$ \u0111\u1ec1u n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n $(O)$. Ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $ABC$ c\u1eaft ti\u1ebfp tuy\u1ebfn c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n k\u1ebb t\u1eeb $C$ t\u1ea1i \u0111i\u1ec3m $D$. S\u1ed1 \u0111o $\\widehat{BDC}$ l\u00e0: ","select":["A. ${{60}^{o}}$ ","B. ${{50}^{o}}$ ","C. ${{40}^{o}}$ ","D. ${{30}^{o}}$"],"explain":"<span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai16/lv1/img\/h934_D16.png' \/><\/center><br\/> Ta c\u00f3: $\\widehat{OBC}=\\dfrac{1}{2}\\widehat{ABC}={{30}^{o}}$ (t\u00ednh ch\u1ea5t ph\u00e2n gi\u00e1c) <br\/> $OB=OC$ (c\u00f9ng b\u1eb1ng b\u00e1n k\u00ednh $(O)$) <br\/> $\\Rightarrow \\Delta OBC$ c\u00e2n t\u1ea1i $O$ <br\/> $\\Rightarrow \\widehat{OBC}=\\widehat{OCB}={{30}^{o}}$ (t\u00ednh ch\u1ea5t tam gi\u00e1c c\u00e2n) <br\/> $\\widehat{COD}=\\widehat{OBC}+\\widehat{OCB}={{60}^{o}}$ (g\u00f3c ngo\u00e0i t\u1ea1i m\u1ed9t \u0111\u1ec9nh c\u1ee7a tam gi\u00e1c) <br\/> M\u00e0 $\\widehat{OCD}={{90}^{o}}$ (t\u00ednh ch\u1ea5t ti\u1ebfp tuy\u1ebfn) <br\/> $\\Rightarrow \\widehat{COD}+\\widehat{CDO}={{90}^{o}}$ (t\u1ed5ng hai g\u00f3c nh\u1ecdn trong tam gi\u00e1c vu\u00f4ng $COD$) <br\/> $\\Rightarrow \\widehat{CDO}={{90}^{o}}-{{60}^{o}}={{30}^{o}}$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D <\/span><\/span>","column":4}]}],"id_ques":1516},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"<span class='basic_left'> Cho n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n $(O)$ \u0111\u01b0\u1eddng k\u00ednh $AB$. Tr\u00ean n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n l\u1ea5y hai \u0111i\u1ec3m $M$ v\u00e0 $N$ sao cho $\\overset\\frown{AM}=\\overset\\frown{MN}=\\overset\\frown{NB}$. T\u1ee9 gi\u00e1c $AMNB$ l\u00e0 h\u00ecnh g\u00ec? ","select":["A. H\u00ecnh ch\u1eef nh\u1eadt ","B. H\u00ecnh thoi ","C. H\u00ecnh thang ","D. H\u00ecnh thang c\u00e2n"],"explain":"<span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai16/lv1/img\/h934_K1.png' \/><\/center><br\/> Ta c\u00f3 $\\widehat{NMB}=\\widehat{MBA}$ (hai g\u00f3c n\u1ed9i ti\u1ebfp ch\u1eafn hai cung b\u1eb1ng nhau) <br\/> $\\Rightarrow MN\/\/AB$ (hai g\u00f3c sole trong b\u1eb1ng nhau) <br\/> $\\Rightarrow AMNB$ l\u00e0 h\u00ecnh thang <br\/> M\u00e0 $\\overset\\frown{AN}=\\overset\\frown{AM}+\\overset\\frown{MN}=\\overset\\frown{BN}+\\overset\\frown{MN}=\\overset\\frown{BM}$ <br\/> $\\Rightarrow \\widehat{MAB}=\\widehat{NBA}$(hai g\u00f3c n\u1ed9i ti\u1ebfp ch\u1eafn hai cung b\u1eb1ng nhau) <br\/> $\\Rightarrow AMNB$ l\u00e0 h\u00ecnh thang c\u00e2n <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D <\/span><\/span>","column":2}]}],"id_ques":1517},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u","temp":"fill_the_blank","correct":[[["60"],["60"]]],"list":[{"point":5,"width":50,"type_input":"","ques":" <span class='basic_left'> Cho \u0111\u01b0\u1eddng tr\u00f2n $(O)$ \u0111\u01b0\u1eddng k\u00ednh $AB$. Tr\u00ean c\u00f9ng m\u1ed9t n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n l\u1ea5y hai \u0111i\u1ec3m $M$ v\u00e0 $N$ sao cho $\\overset\\frown{AM}=\\overset\\frown{MN}=\\overset\\frown{NB}$. G\u1ecdi $P$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $AM$ v\u00e0 $BN$. Ti\u1ebfp tuy\u1ebfn c\u1ee7a n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n t\u1ea1i $B$ v\u00e0 $M$ c\u1eaft nhau \u1edf $Q$. <br\/> <b> C\u00e2u 1: <\/b> Khi \u0111\u00f3: $\\widehat{APB}=$ _input_ $^o$; $\\widehat{BQM}=$ _input_ $^o$","explain":" <span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai16/lv1/img\/h934_D18.png' \/><\/center> <br\/> Ta c\u00f3: $\\overset\\frown{AM}=\\overset\\frown{MN}=\\overset\\frown{NB}$ <br\/> $\\Rightarrow \\text{s\u0111}\\overset\\frown{AM}=\\text{s\u0111}\\overset\\frown{MN}=\\text{s\u0111}\\overset\\frown{NB}=\\dfrac{\\text{s\u0111}\\overset\\frown{AB}}{3}={{60}^{o}}$ <br\/> $\\text{s\u0111}\\overset\\frown{BAM}=\\text{s\u0111}\\overset\\frown{AB}+\\text{s\u0111}\\overset\\frown{AM}={{180}^{o}}+{{60}^{o}}={{240}^{o}}$ <br\/> $\\text{s\u0111}\\overset\\frown{MNB}=\\text{s\u0111}\\overset\\frown{MN}+\\text{s\u0111}\\overset\\frown{NB}={{60}^{o}}+{{60}^{o}}={{120}^{o}}$ <br\/> $\\widehat{APB}=\\dfrac{\\text{s\u0111}\\overset\\frown{AB}-\\text{s\u0111}\\overset\\frown{MN}}{2}$ (\u0111\u1ecbnh l\u00ed g\u00f3c c\u00f3 \u0111\u1ec9nh n\u1eb1m \u1edf ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n) <br\/> $\\Rightarrow \\widehat{APB}=\\dfrac{{{180}^{o}}-{{60}^{o}}}{2}={{60}^{o}}$ <br\/> $\\widehat{BQM}=\\dfrac{\\text{s\u0111}\\overset\\frown{BAM}-\\text{s\u0111}\\overset\\frown{MNB}}{2}$ (\u0111\u1ecbnh l\u00ed g\u00f3c c\u00f3 \u0111\u1ec9nh n\u1eb1m \u1edf ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n) <br\/> $\\Rightarrow \\widehat{BQM}=\\dfrac{{{240}^{o}}-{{120}^{o}}}{2}={{60}^{o}}$ <br\/> <span class='basic_pink'> V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u1ea7n l\u01b0\u1ee3t l\u00e0 $60$ v\u00e0 $60$ <\/span><\/span> "}]}],"id_ques":1518},{"time":24,"part":[{"time":3,"title":"<span class='basic_left'> Cho \u0111\u01b0\u1eddng tr\u00f2n $(O)$ \u0111\u01b0\u1eddng k\u00ednh $AB$. Tr\u00ean c\u00f9ng m\u1ed9t n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n l\u1ea5y hai \u0111i\u1ec3m $M$ v\u00e0 $N$ sao cho $\\overset\\frown{AM}=\\overset\\frown{MN}=\\overset\\frown{NB}$. G\u1ecdi $P$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $AM$ v\u00e0 $BN$. Ti\u1ebfp tuy\u1ebfn c\u1ee7a n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n t\u1ea1i $B$ v\u00e0 $M$ c\u1eaft nhau \u1edf $Q$. <br\/> <b> C\u00e2u 2: <\/b> Ch\u1ee9ng minh $MN$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $QMB$","title_trans":"S\u1eafp x\u1ebfp c\u00e1c c\u00e2u \u0111\u1ec3 \u0111\u01b0\u1ee3c b\u00e0i ch\u1ee9ng minh","temp":"sequence","correct":[[[3],[2],[1]]],"list":[{"point":5,"left":["$\\Rightarrow MN$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $QMB$ ","$\\Rightarrow \\widehat{QMN}=\\widehat{BMN}$ (h\u1ec7 qu\u1ea3 g\u00f3c n\u1ed9i ti\u1ebfp) ","Ta c\u00f3: $ \\text{s\u0111}\\overset\\frown{MN}=\\text{s\u0111}\\overset\\frown{NB}={{60}^{o}}$ (theo c\u00e2u 1)"],"top":75,"explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai16/lv1/img\/h934_D19.png' \/><\/center> <br\/> Ta c\u00f3: $ \\text{s\u0111}\\overset\\frown{MN}=\\text{s\u0111}\\overset\\frown{NB}={{60}^{o}}$ (theo c\u00e2u 1) <br\/> $\\Rightarrow \\widehat{QMN}=\\widehat{BMN}$ (h\u1ec7 qu\u1ea3 g\u00f3c n\u1ed9i ti\u1ebfp) <br\/> $\\Rightarrow MN$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $QMB$ <\/span>"}]}],"id_ques":1519},{"time":24,"part":[{"time":3,"title":"<span class='basic_left'> Cho tam gi\u00e1c $ABC$ n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n $(O)$. Tr\u00ean cung nh\u1ecf $AC$ l\u1ea5y \u0111i\u1ec3m $M$, tia $AM$ v\u00e0 tia $BC$ c\u1eaft nhau t\u1ea1i $S$. G\u1ecdi $I$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $AC$ v\u00e0 $BM$. Ch\u1ee9ng minh $\\widehat{MIC}+\\widehat{ASB}=2\\widehat{ACB}$. ","title_trans":"S\u1eafp x\u1ebfp c\u00e1c c\u00e2u \u0111\u1ec3 \u0111\u01b0\u1ee3c b\u00e0i ch\u1ee9ng minh","temp":"sequence","correct":[[[3],[5],[1],[4],[2]]],"list":[{"point":5,"left":["$\\Rightarrow \\widehat{MIC}+\\widehat{ASB}=\\dfrac{\\text{s\u0111}\\overset\\frown{AB}+\\text{s\u0111}\\overset\\frown{MC}}{2}+\\dfrac{\\text{s\u0111}\\overset\\frown{AB}-\\text{s\u0111}\\overset\\frown{MC}}{2} =\\text{s\u0111}\\overset\\frown{AB}\\,(1)$ ","T\u1eeb (1) v\u00e0 (2) $\\Rightarrow \\widehat{MIC}+\\widehat{ASB}=2\\widehat{ACB}$. "," Ta c\u00f3: $\\widehat{MIC}=\\dfrac{\\text{s\u0111}\\overset\\frown{AB}+\\text{s\u0111}\\overset\\frown{MC}}{2}$ (\u0111\u1ecbnh l\u00ed g\u00f3c c\u00f3 \u0111\u1ec9nh n\u1eb1m \u1edf trong \u0111\u01b0\u1eddng tr\u00f2n)"," M\u00e0 $\\text{s\u0111}\\overset\\frown{AB}=2\\widehat{ACB}\\,(2)$ (\u0111\u1ecbnh l\u00ed g\u00f3c n\u1ed9i ti\u1ebfp)","$\\widehat{ASB}=\\dfrac{\\text{s\u0111}\\overset\\frown{AB}-\\text{s\u0111}\\overset\\frown{MC}}{2}$ (\u0111\u1ecbnh l\u00ed g\u00f3c c\u00f3 \u0111\u1ec9nh n\u1eb1m \u1edf ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n)"],"top":75,"explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai16/lv1/img\/h934_D20.png' \/><\/center> <br\/>Ta c\u00f3: $\\widehat{MIC}=\\dfrac{\\text{s\u0111}\\overset\\frown{AB}+\\text{s\u0111}\\overset\\frown{MC}}{2}$ (\u0111\u1ecbnh l\u00ed g\u00f3c c\u00f3 \u0111\u1ec9nh n\u1eb1m \u1edf trong \u0111\u01b0\u1eddng tr\u00f2n) <br\/> $\\widehat{ASB}=\\dfrac{\\text{s\u0111}\\overset\\frown{AB}-\\text{s\u0111}\\overset\\frown{MC}}{2}$ (\u0111\u1ecbnh l\u00ed g\u00f3c c\u00f3 \u0111\u1ec9nh n\u1eb1m \u1edf ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n) <br\/> $\\Rightarrow \\widehat{MIC}+\\widehat{ASB}=\\dfrac{\\text{s\u0111}\\overset\\frown{AB}+\\text{s\u0111}\\overset\\frown{MC}}{2}+\\dfrac{\\text{s\u0111}\\overset\\frown{AB}-\\text{s\u0111}\\overset\\frown{MC}}{2} =\\text{s\u0111}\\overset\\frown{AB}\\,(1)$ <br\/> M\u00e0 $\\text{s\u0111}\\overset\\frown{AB}=2\\widehat{ACB}\\,(2)$ (\u0111\u1ecbnh l\u00ed g\u00f3c n\u1ed9i ti\u1ebfp) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow \\widehat{MIC}+\\widehat{ASB}=2\\widehat{ACB}$. <\/span>"}]}],"id_ques":1520}],"lesson":{"save":0,"level":1}}