{"segment":[{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"<span class='basic_left'> Cho tam gi\u00e1c $ABC$ c\u00e2n t\u1ea1i $A$ n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n $(O)$. Ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $C$ v\u00e0 $B$ c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n l\u1ea7n l\u01b0\u1ee3t t\u1ea1i $D$ v\u00e0 $F$. G\u1ecdi $E$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $CD$ v\u00e0 $BF$. T\u1ee9 gi\u00e1c $ADEF$ l\u00e0 h\u00ecnh g\u00ec? ","select":["A. H\u00ecnh ch\u1eef nh\u1eadt ","B. H\u00ecnh thoi ","C. H\u00ecnh vu\u00f4ng ","D. H\u00ecnh b\u00ecnh h\u00e0nh"],"explain":"<span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai16/lv3/img\/h934_D17.png' \/><\/center><br\/> Ta c\u00f3: $AB = AC \\Rightarrow \\overset\\frown{ADB}=\\overset\\frown{AFC}$ (\u0111\u1ecbnh l\u00ed li\u00ean h\u1ec7 gi\u1eefa cung v\u00e0 d\u00e2y)<br\/> $\\widehat{ABF}=\\widehat{CBF};\\,\\widehat{ACD}=\\widehat{BCD}$ (t\u00ednh ch\u1ea5t ph\u00e2n gi\u00e1c) <br\/> $\\Rightarrow \\overset\\frown{AF}=\\overset\\frown{FC}=\\dfrac{\\overset\\frown{AC}}{2};\\,\\overset\\frown{AD}=\\overset\\frown{DB}=\\dfrac{\\overset\\frown{AB}}{2}$ (h\u1ec7 qu\u1ea3 g\u00f3c n\u1ed9i ti\u1ebfp) <br\/> $\\Rightarrow \\overset\\frown{AD}=\\overset\\frown{DB}=\\overset\\frown{FC}=\\overset\\frown{AF}$ <br\/> $\\Rightarrow AD = DB = AF = FC \\,(1)$ (\u0111\u1ecbnh l\u00ed li\u00ean h\u1ec7 gi\u1eefa cung v\u00e0 d\u00e2y) <br\/> L\u1ea1i c\u00f3: $\\widehat{DBF}=\\dfrac{1}{2}\\text{s\u0111}\\overset\\frown{DF}$ (\u0111\u1ecbnh l\u00ed g\u00f3c n\u1ed9i ti\u1ebfp) <br\/> $\\Rightarrow \\widehat{DBF}=\\dfrac{\\text{s\u0111}\\overset\\frown{DA}+\\text{s\u0111}\\overset\\frown{AF}}{2}=\\dfrac{\\text{s\u0111}\\overset\\frown{DB}+\\text{s\u0111}\\overset\\frown{FC}}{2}=\\widehat{DEB}$ (\u0111\u1ecbnh l\u00ed g\u00f3c c\u00f3 \u0111\u1ec9nh n\u1eb1m trong \u0111\u01b0\u1eddng tr\u00f2n) <br\/> $\\Rightarrow \\Delta BDE$ c\u00e2n t\u1ea1i $D$ (d\u1ea5u hi\u1ec7u nh\u1eadn bi\u1ebft) <br\/> $\\Rightarrow DB=DE\\,(2)$ <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow DA=DE$ <br\/> Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1: $AF=FE$ <br\/> $\\Rightarrow AD=DE=EF=FA$ <br\/> $\\Rightarrow ADEF$ l\u00e0 h\u00ecnh thoi <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B <\/span><\/span>","column":4}]}],"id_ques":1521},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["90"]]],"list":[{"point":10,"width":50,"type_input":"","ques":" <span class='basic_left'>Tr\u00ean \u0111\u01b0\u1eddng tr\u00f2n $(O)$ cho c\u00e1c \u0111i\u1ec3m $A, B, C, D$ theo th\u1ee9 t\u1ef1 \u0111\u00f3. G\u1ecdi $M, N, P, Q$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 c\u00e1c \u0111i\u1ec3m ch\u00ednh gi\u1eefa c\u1ee7a c\u00e1c cung $AB, BC, CD, DA$; $I$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $MP$ v\u00e0 $NQ$. S\u1ed1 \u0111o g\u00f3c $MIN$ l\u00e0 _input_ $^o$ ","explain":"<span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai16/lv3/img\/h934_K2.png' \/><\/center> <br\/>Ta c\u00f3: $\\overset\\frown{AM}=\\overset\\frown{MB} = \\dfrac{1}{2}\\overset\\frown{AB}\\,\\overset\\frown{BN}=\\overset\\frown{NC} = \\dfrac{1}{2}\\overset\\frown{BC};\\,\\overset\\frown{CP}=\\overset\\frown{PD} = \\dfrac{1}{2}\\overset\\frown{CD};\\,\\overset\\frown{DQ}=\\overset\\frown{QA} = \\dfrac{1}{2}\\overset\\frown{DA}$ (gi\u1ea3 thi\u1ebft) <br\/> $\\text{s\u0111}\\overset\\frown{AMB}+\\text{s\u0111}\\overset\\frown{BNC}+\\text{s\u0111}\\overset\\frown{CPD}+\\text{s\u0111}\\overset\\frown{DQA}={{360}^{o}}$ (s\u1ed1 \u0111o c\u1ee7a cung c\u1ea3 \u0111\u01b0\u1eddng tr\u00f2n) <br\/> $\\widehat{MIN}=\\dfrac{\\text{s\u0111}\\overset\\frown{MBN}+\\text{s\u0111}\\overset\\frown{PDQ}}{2}$ (\u0111\u1ecbnh l\u00ed g\u00f3c c\u00f3 \u0111\u1ec9nh n\u1eb1m trong \u0111\u01b0\u1eddng tr\u00f2n) <br\/> $\\Rightarrow \\widehat{MIN}=\\dfrac{1}{2}\\left( \\text{s\u0111}\\overset\\frown{MB}+\\text{s\u0111}\\overset\\frown{BN}+\\text{s\u0111}\\overset\\frown{PD}+\\text{s\u0111}\\overset\\frown{DQ} \\right) \\\\ =\\dfrac{1}{4}\\left( \\text{s\u0111}\\overset\\frown{AB}+\\text{s\u0111}\\overset\\frown{BC}+\\text{s\u0111}\\overset\\frown{CD}+\\text{s\u0111}\\overset\\frown{DA} \\right)\\\\ =\\dfrac{1}{4}{{.360}^{o}}={{90}^{o}}$ <br\/> <span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $90$<\/span> <\/span>"}]}],"id_ques":1522},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"<span class='basic_left'>Cho \u0111\u01b0\u1eddng tr\u00f2n $(O;\\,R)$ \u0111\u01b0\u1eddng k\u00ednh $BC$. T\u1eeb m\u1ed9t \u0111i\u1ec3m $A$ tr\u00ean tia $BC$ v\u1ebd c\u00e1t tuy\u1ebfn $AED$ v\u1edbi \u0111\u01b0\u1eddng tr\u00f2n sao cho $AE=R$ v\u00e0 $\\widehat{BOD}={{45}^{o}}.$ S\u1ed1 \u0111o g\u00f3c $EAO$ l\u00e0: ","select":["A. ${{15}^{o}}$ ","B. $22,{{5}^{o}}$ ","C. ${{45}^{o}}$ ","D. $67,{{5}^{o}}$"],"explain":"<span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai16/lv3/img\/h934_K3.png' \/><\/center><br\/> Ta c\u00f3: $AE=OE=R$ <br\/> $\\Rightarrow \\Delta AOE$ c\u00e2n t\u1ea1i $E$ <br\/> $\\Rightarrow \\widehat{EAO}=\\widehat{EOA}$ (t\u00ednh ch\u1ea5t tam gi\u00e1c c\u00e2n) <br\/> Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1, ta \u0111\u01b0\u1ee3c: $ \\widehat{OED}=\\widehat{ODE}$ <br\/> Ta c\u00f3: $\\widehat{OED}=\\widehat{EAO}+\\widehat{AOE}$ (g\u00f3c ngo\u00e0i t\u1ea1i m\u1ed9t \u0111\u1ec9nh c\u1ee7a tam gi\u00e1c $AEO$) <br\/> $\\Rightarrow \\widehat{OED}=\\widehat{ODE}=2\\widehat{EAO}$ <br\/> M\u1eb7t kh\u00e1c $\\widehat{DOB}=\\widehat{DAO}+\\widehat{ADO}$ (g\u00f3c ngo\u00e0i t\u1ea1i m\u1ed9t \u0111\u1ec9nh c\u1ee7a tam gi\u00e1c $ADO$) <br\/> $\\Rightarrow \\widehat{DOB}=3\\widehat{EAO}$ <br\/> $\\Rightarrow \\widehat{EAO}=\\dfrac{\\widehat{DOB}}{3}={{15}^{o}}$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A <\/span><\/span>","column":4}]}],"id_ques":1523},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["120"]]],"list":[{"point":10,"width":50,"type_input":"","ques":" <span class='basic_left'>Tr\u00ean n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n $(O; R)$, \u0111\u01b0\u1eddng k\u00ednh $AB$ l\u1ea5y l\u1ea7n l\u01b0\u1ee3t hai \u0111i\u1ec3m $C, D$ sao cho $C$ n\u1eb1m gi\u1eefa $A$ v\u00e0 $D$. G\u1ecdi $E$ v\u00e0 $F$ theo th\u1ee9 t\u1ef1 l\u00e0 h\u00ecnh chi\u1ebfu vu\u00f4ng g\u00f3c c\u1ee7a $A, B$ tr\u00ean \u0111\u01b0\u1eddng th\u1eb3ng $CD$. Tia $AD$ c\u1eaft $BC$ t\u1ea1i $I$. Bi\u1ebft r\u1eb1ng $AE+BF=R\\sqrt{3}$. S\u1ed1 \u0111o g\u00f3c $AIB$ l\u00e0 _input_ $^o$ ","explain":"<span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai16/lv3/img\/h934_K4.png' \/><\/center> <br\/> K\u1ebb $OH\\bot CD \\Rightarrow DH=HC$ (\u0111\u1ecbnh l\u00ed \u0111\u01b0\u1eddng k\u00ednh v\u00e0 d\u00e2y cung) <br\/> Ta c\u00f3: $\\left\\{ \\begin{align} & AE\\bot CD \\\\ & BF\\bot CD \\\\ \\end{align} \\right.$ (gi\u1ea3 thi\u1ebft) <br\/> $\\Rightarrow AE\/\/BF$ (t\u1eeb vu\u00f4ng g\u00f3c \u0111\u1ebfn song song) <br\/> $\\Rightarrow AEFB$ l\u00e0 h\u00ecnh thang <br\/> M\u00e0 $\\left\\{ \\begin{align} & OH\\bot CD\\Rightarrow OH\/\/AE\/\/BF \\\\ & OA=OB=R \\\\ \\end{align} \\right.$ <br\/> $\\Rightarrow OH$ l\u00e0 \u0111\u01b0\u1eddng trung b\u00ecnh c\u1ee7a h\u00ecnh thang $AEFB$ <br\/> $\\Rightarrow OH=\\dfrac{AE+BF}{2}=\\dfrac{R\\sqrt{3}}{2}$ <br\/> X\u00e9t $\\Delta OHD$ vu\u00f4ng t\u1ea1i $H$ c\u00f3: <br\/> $DH=\\sqrt{O{{D}^{2}}-O{{H}^{2}}}=\\sqrt{{{R}^{2}}-\\dfrac{3{{R}^{2}}}{4}}=\\dfrac{R}{2}$ <br\/> $\\Rightarrow DC=2DH=R$ <br\/> $\\Rightarrow OD=OC=DC=R$ <br\/> $\\Rightarrow \\Delta ODC$ \u0111\u1ec1u (\u0111\u1ecbnh ngh\u0129a tam gi\u00e1c \u0111\u1ec1u) <br\/> $\\Rightarrow \\widehat{DOC}=\\text{s\u0111}\\overset\\frown{DC}={{60}^{o}}$ (t\u00ednh ch\u1ea5t tam gi\u00e1c \u0111\u1ec1u v\u00e0 \u0111\u1ecbnh ngh\u0129a s\u1ed1 \u0111o cung) <br\/> $\\Rightarrow \\widehat{AIB}=\\dfrac{\\text{s\u0111}\\overset\\frown{AB}+\\text{s\u0111}\\overset\\frown{CD}}{2}$ (\u0111\u1ecbnh l\u00ed g\u00f3c c\u00f3 \u0111\u1ec9nh n\u1eb1m trong \u0111\u01b0\u1eddng tr\u00f2n) <br\/> $\\Rightarrow \\widehat{AIB}=\\dfrac{{{180}^{o}}+{{60}^{o}}}{2}={{120}^{o}}$ <br\/> <span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $120$<\/span> <\/span>"}]}],"id_ques":1524},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"<span class='basic_left'> Cho tam gi\u00e1c $ABC$ n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n $(O)$, c\u00e1c tia ph\u00e2n gi\u00e1c c\u1ee7a c\u00e1c g\u00f3c $A$ v\u00e0 $B$ c\u1eaft nhau t\u1ea1i $I$ v\u00e0 c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n theo th\u1ee9 t\u1ef1 t\u1ea1i $D$ v\u00e0 $E$. Khi \u0111\u00f3: ","select":["A. $DE$ song song v\u1edbi $IC$ ","B. $DE$ \u0111i qua trung \u0111i\u1ec3m c\u1ee7a $IC$ ","C. $DE$ vu\u00f4ng g\u00f3c v\u1edbi $IC$ ","D. $DE$ l\u00e0 trung tr\u1ef1c c\u1ee7a $IC$"],"explain":"<span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai16/lv3/img\/h934_K5.png' \/><\/center><br\/> Ta c\u00f3: $\\widehat{ABE}=\\widehat{CBE}$ (t\u00ednh ch\u1ea5t ph\u00e2n gi\u00e1c) <br\/> $\\Rightarrow \\overset\\frown{AE}=\\overset\\frown{EC}$ (h\u1ec7 qu\u1ea3 g\u00f3c n\u1ed9i ti\u1ebfp) <br\/> Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1: $\\overset\\frown{BD}=\\overset\\frown{DC}$ <br\/> $\\Rightarrow BD=DC$ (\u0111\u1ecbnh l\u00ed li\u00ean h\u1ec7 gi\u1eefa cung v\u00e0 d\u00e2y) (1) <br\/> M\u00e0 $\\widehat{EBD}=\\dfrac{1}{2}\\text{s\u0111}\\overset\\frown{DCE}$ (\u0111\u1ecbnh l\u00ed g\u00f3c n\u1ed9i ti\u1ebfp) <br\/> $\\Rightarrow \\widehat{EBD}=\\dfrac{1}{2}\\text{s\u0111}\\overset\\frown{DCE}=\\dfrac{1}{2}\\left( \\text{s\u0111}\\overset\\frown{DC}+\\text{s\u0111}\\overset\\frown{EC} \\right)=\\dfrac{1}{2}\\left( \\text{s\u0111}\\overset\\frown{BD}+\\text{s\u0111}\\overset\\frown{AE} \\right)=\\widehat{BID}$ <br\/> $\\Rightarrow \\Delta BID$ c\u00e2n t\u1ea1i $D$ <br\/> $\\Rightarrow DB=DI$(\u0111\u1ecbnh ngh\u0129a tam gi\u00e1c c\u00e2n) (2) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow DI=DC$ <br\/> Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1, ta \u0111\u01b0\u1ee3c: $EI=EC$ <br\/> $\\Rightarrow DE$ l\u00e0 trung tr\u1ef1c c\u1ee7a $IC$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D <\/span><\/span>","column":2}]}],"id_ques":1525},{"time":24,"part":[{"time":3,"title":"Cho tam gi\u00e1c $ABC$ c\u00e2n t\u1ea1i $A$ n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n $(O)$. L\u1ea5y \u0111i\u1ec3m $M$ thu\u1ed9c tia \u0111\u1ed1i c\u1ee7a tia $BC$. G\u1ecdi $I$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $MA$ v\u1edbi \u0111\u01b0\u1eddng tr\u00f2n. Ch\u1ee9ng minh $AI.AM=AC^2$.","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],[6],[4],[2]]],"list":[{"point":10,"left":[" $\\Rightarrow \\widehat{AMC}=\\dfrac{\\text{s\u0111}\\overset\\frown{AIB}-\\text{s\u0111}\\overset\\frown{IB}}{2}=\\dfrac{\\text{s\u0111}\\overset\\frown{IA}}{2}=\\widehat{ACI}$ ","Ta c\u00f3: $AB=AC$ ($\\Delta ABC$ c\u00e2n t\u1ea1i $A$) <br\/> $\\Rightarrow \\overset\\frown{AIB}=\\overset\\frown{AC}$ (\u0111\u1ecbnh l\u00ed li\u00ean h\u1ec7 gi\u1eefa cung v\u00e0 d\u00e2y) "," $\\Rightarrow \\Delta AIC\\sim \\Delta ACM$(g.g) $\\Rightarrow \\dfrac{AI}{AC}=\\dfrac{AC}{AM}$(t\u1ec9 s\u1ed1 \u0111\u1ed3ng d\u1ea1ng)","$\\Rightarrow AI.AM=A{{C}^{2}}$"," X\u00e9t $\\Delta AIC$ v\u00e0 $\\Delta ACM$ c\u00f3: $\\left\\{ \\begin{align} & \\widehat{A} \\,\\text{chung} \\\\ & \\widehat{AMC}=\\widehat{ACI}\\,\\left( \\text{ch\u1ee9ng minh tr\u00ean} \\right) \\\\ \\end{align} \\right.$ ","$\\widehat{AMC}=\\dfrac{\\text{s\u0111}\\overset\\frown{AC}-\\text{s\u0111}\\overset\\frown{IB}}{2}$ (\u0111\u1ecbnh l\u00ed g\u00f3c c\u00f3 \u0111\u1ec9nh n\u1eb1m \u1edf ngo\u00e0i \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/lv3/img\/h934_K6.png' \/><\/center> <br\/>Ta c\u00f3: $AB=AC$ ($\\Delta ABC$ c\u00e2n t\u1ea1i $A$) <br\/> $\\Rightarrow \\overset\\frown{AIB}=\\overset\\frown{AC}$ (\u0111\u1ecbnh l\u00ed li\u00ean h\u1ec7 gi\u1eefa cung v\u00e0 d\u00e2y) <br\/> $\\widehat{AMC}=\\dfrac{\\text{s\u0111}\\overset\\frown{AC}-\\text{s\u0111}\\overset\\frown{IB}}{2}$ (\u0111\u1ecbnh l\u00ed g\u00f3c c\u00f3 \u0111\u1ec9nh n\u1eb1m \u1edf ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n) <br\/> $\\Rightarrow \\widehat{AMC}=\\dfrac{\\text{s\u0111}\\overset\\frown{AIB}-\\text{s\u0111}\\overset\\frown{IB}}{2}=\\dfrac{\\text{s\u0111}\\overset\\frown{IA}}{2}=\\widehat{ACI}$ <br\/> X\u00e9t $\\Delta AIC$ v\u00e0 $\\Delta ACM$ c\u00f3: <br\/> $\\left\\{ \\begin{align} & \\widehat{A} \\, \\text{chung} \\\\ & \\widehat{AMC}=\\widehat{ACI}\\,\\left( \\text{ch\u1ee9ng minh tr\u00ean} \\right) \\\\ \\end{align} \\right.$ <br\/> $\\Rightarrow \\Delta AIC\\backsim \\Delta ACM$(g.g) <br\/> $\\Rightarrow \\dfrac{AI}{AC}=\\dfrac{AC}{AM}$(t\u1ec9 s\u1ed1 \u0111\u1ed3ng d\u1ea1ng) <br\/> $\\Rightarrow AI.AM=A{{C}^{2}}$ <\/span>"}]}],"id_ques":1526},{"time":24,"part":[{"time":3,"title":"Cho \u0111\u01b0\u1eddng tr\u00f2n $(O;R)$, hai \u0111\u01b0\u1eddng k\u00ednh $AB$ v\u00e0 $CD$ vu\u00f4ng g\u00f3c v\u1edbi nhau. Tr\u00ean \u0111\u01b0\u1eddng k\u00ednh $AB$ l\u1ea5y \u0111i\u1ec3m $E$ sao cho $AE=R\\sqrt{2}$. Qua $E$ v\u1ebd d\u00e2y $CF$. Ti\u1ebfp tuy\u1ebfn c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n t\u1ea1i $F$ c\u1eaft \u0111\u01b0\u1eddng th\u1eb3ng $CD$ t\u1ea1i $M$. <br\/> <b> C\u00e2u 1: <\/b> Ch\u1ee9ng minh $MF \/\/ AC$. ","title_trans":"S\u1eafp x\u1ebfp c\u00e1c c\u00e2u \u0111\u1ec3 \u0111\u01b0\u1ee3c b\u00e0i ch\u1ee9ng minh","temp":"sequence","correct":[[[5],[1],[6],[3],[2],[4]]],"list":[{"point":10,"left":[" T\u1eeb (1) v\u00e0 (2) $\\Rightarrow \\widehat{CFx}=\\widehat{ACE}$","X\u00e9t $\\Delta OAC$ vu\u00f4ng t\u1ea1i $O$ c\u00f3: $AC=\\sqrt{O{{A}^{2}}+O{{C}^{2}}}=\\sqrt{{{R}^{2}}+{{R}^{2}}}=R\\sqrt{2}$"," $\\Rightarrow MF\/\/AC$ (hai g\u00f3c so le trong b\u1eb1ng nhau) ","Ta c\u00f3: $\\widehat{CFx}=\\dfrac{1}{2}\\text{s\u0111}\\overset\\frown{FC}$ (g\u00f3c t\u1ea1o b\u1edfi tia ti\u1ebfp tuy\u1ebfn v\u00e0 d\u00e2y cung ch\u1eafn cung $FBC$)"," $\\Rightarrow AC=AE=R\\sqrt{2}$ $\\Rightarrow \\Delta ACE$ c\u00e2n t\u1ea1i $A$ <br\/> $\\Rightarrow \\widehat{ACE}=\\widehat{AEC}\\,(1)$ (t\u00ednh ch\u1ea5t tam gi\u00e1c c\u00e2n) "," $\\Rightarrow \\widehat{CFx}=\\dfrac{\\text{s\u0111}\\overset\\frown{FB}+\\text{s\u0111}\\overset\\frown{BC}}{2}=\\dfrac{\\text{s\u0111}\\overset\\frown{FB}+\\text{s\u0111}\\overset\\frown{AC}}{2}=\\widehat{AEC}\\,(2)$ "],"top":100,"explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai16/lv3/img\/h934_K7.png' \/><\/center> <br\/> X\u00e9t $\\Delta OAC$ vu\u00f4ng t\u1ea1i $O$ c\u00f3: <br\/> $AC=\\sqrt{O{{A}^{2}}+O{{C}^{2}}}=\\sqrt{{{R}^{2}}+{{R}^{2}}}=R\\sqrt{2}$ <br\/> $\\Rightarrow AC=AE=R\\sqrt{2}$ <br\/> $\\Rightarrow \\Delta ACE$ c\u00e2n t\u1ea1i $A$ <br\/> $\\Rightarrow \\widehat{ACE}=\\widehat{AEC}\\,(1)$ (t\u00ednh ch\u1ea5t tam gi\u00e1c c\u00e2n) <br\/> Ta c\u00f3: $\\widehat{CFx}=\\dfrac{1}{2}\\text{s\u0111}\\overset\\frown{FC}$ (g\u00f3c t\u1ea1o b\u1edfi tia ti\u1ebfp tuy\u1ebfn v\u00e0 d\u00e2y cung ch\u1eafn cung $FBC$) <br\/> $\\Rightarrow \\widehat{CFx}=\\dfrac{\\text{s\u0111}\\overset\\frown{FB}+\\text{s\u0111}\\overset\\frown{BC}}{2}=\\dfrac{\\text{s\u0111}\\overset\\frown{FB}+\\text{s\u0111}\\overset\\frown{AC}}{2}=\\widehat{AEC}\\,(2)$ <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow \\widehat{CFx}=\\widehat{ACE}$ <br\/> $\\Rightarrow MF\/\/AC$ (hai g\u00f3c so le trong b\u1eb1ng nhau) <\/span>"}]}],"id_ques":1527},{"time":24,"part":[{"time":3,"title":"Cho \u0111\u01b0\u1eddng tr\u00f2n $(O;R)$, hai \u0111\u01b0\u1eddng k\u00ednh $AB$ v\u00e0 $CD$ vu\u00f4ng g\u00f3c v\u1edbi nhau. Tr\u00ean \u0111\u01b0\u1eddng k\u00ednh $AB$ l\u1ea5y \u0111i\u1ec3m $E$ sao cho $AE=R\\sqrt{2}$. Qua $E$ v\u1ebd d\u00e2y $CF$. Ti\u1ebfp tuy\u1ebfn c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n t\u1ea1i $F$ c\u1eaft \u0111\u01b0\u1eddng th\u1eb3ng $CD$ t\u1ea1i $M$. <br\/> <b> C\u00e2u 2: <\/b> Ch\u1ee9ng minh $CF$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $BCD$. ","title_trans":"S\u1eafp x\u1ebfp c\u00e1c c\u00e2u \u0111\u1ec3 \u0111\u01b0\u1ee3c b\u00e0i ch\u1ee9ng minh","temp":"sequence","correct":[[[2],[4],[3],[1]]],"list":[{"point":10,"left":[" $\\Rightarrow \\text{s\u0111}\\overset\\frown{ADF}=2\\widehat{MFA}=2\\widehat{FAC}=\\text{s\u0111}\\overset\\frown{FBC}$ (\u0111\u1ecbnh l\u00ed g\u00f3c n\u1ed9i ti\u1ebfp v\u00e0 g\u00f3c t\u1ea1o b\u1edfi tia ti\u1ebfp tuy\u1ebfn v\u00e0 d\u00e2y cung) ","$\\Rightarrow \\widehat{DCF}=\\widehat{BCF}$ (h\u1ec7 qu\u1ea3 g\u00f3c n\u1ed9i ti\u1ebfp) <br\/> $\\Rightarrow CF$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $BCD$"," M\u00e0 $\\text{s\u0111}\\overset\\frown{AD}=\\text{s\u0111}\\overset\\frown{BC}={{90}^{o}}$ $\\Rightarrow \\overset\\frown{DF}=\\overset\\frown{BF}$"," Ta c\u00f3: $MF \/\/AC$ (theo c\u00e2u 1) <br\/> $\\Rightarrow \\widehat{MFA}=\\widehat{FAC}$ (hai g\u00f3c sole trong) "],"top":100,"explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai16/lv3/img\/h934_K8.png' \/><\/center> <br\/> Ta c\u00f3: $MF \/\/AC$ (theo c\u00e2u 1) <br\/> $\\Rightarrow \\widehat{MFA}=\\widehat{FAC}$ (hai g\u00f3c sole trong) <br\/> $\\Rightarrow \\text{s\u0111}\\overset\\frown{ADF}=2\\widehat{MFA}=2\\widehat{FAC}=\\text{s\u0111}\\overset\\frown{FBC}$ (\u0111\u1ecbnh l\u00ed g\u00f3c n\u1ed9i ti\u1ebfp v\u00e0 g\u00f3c t\u1ea1o b\u1edfi tia ti\u1ebfp tuy\u1ebfn v\u00e0 d\u00e2y cung) <br\/> M\u00e0 $\\text{s\u0111}\\overset\\frown{AD}=\\text{s\u0111}\\overset\\frown{BC}={{90}^{o}}$ $\\Rightarrow \\overset\\frown{DF}=\\overset\\frown{BF}$ <br\/> $\\Rightarrow \\widehat{DCF}=\\widehat{BCF}$ (h\u1ec7 qu\u1ea3 g\u00f3c n\u1ed9i ti\u1ebfp) <br\/> $\\Rightarrow CF$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $BCD$ <\/span>"}]}],"id_ques":1528},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"<span class='basic_left'> Cho \u0111\u01b0\u1eddng tr\u00f2n $(O)$, d\u00e2y cung $AB$. G\u1ecdi $I$ l\u00e0 \u0111i\u1ec3m thu\u1ed9c d\u00e2y $AB$ ($IA > IB$), $D$ l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa c\u1ee7a cung $AB$. Tia $DI$ c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n \u1edf $C$, ti\u1ebfp tuy\u1ebfn c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n t\u1ea1i $C$ c\u1eaft $AB$ t\u1ea1i $K$. <br\/> <b> C\u00e2u 1:<\/b> Tam gi\u00e1c $KIC$ l\u00e0 tam gi\u00e1c g\u00ec? ","select":["A. Tam gi\u00e1c c\u00e2n ","B. Tam gi\u00e1c \u0111\u1ec1u ","C. Tam gi\u00e1c vu\u00f4ng ","D. Tam gi\u00e1c vu\u00f4ng c\u00e2n"],"explain":"<span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai16/lv3/img\/h934_K9.png' \/><\/center><br\/> Ta c\u00f3: $\\overset\\frown{AD}=\\overset\\frown{DB}$ (gi\u1ea3 thi\u1ebft) <br\/> $\\widehat{KCD}=\\dfrac{\\text{s\u0111}\\overset\\frown{CBD}}{2}$ (\u0111\u1ecbnh l\u00ed g\u00f3c t\u1ea1o b\u1edfi tia ti\u1ebfp tuy\u1ebfn v\u00e0 d\u00e2y cung) <br\/> $\\Rightarrow \\widehat{KCD}=\\dfrac{\\text{s\u0111}\\overset\\frown{BC}+\\text{s\u0111}\\overset\\frown{BD}}{2}=\\dfrac{\\text{s\u0111}\\overset\\frown{BC}+\\text{s\u0111}\\overset\\frown{AD}}{2}=\\widehat{CIK}$ (\u0111\u1ecbnh l\u00ed g\u00f3c c\u00f3 \u0111\u1ec9nh n\u1eb1m trong \u0111\u01b0\u1eddng tr\u00f2n) <br\/> $\\Rightarrow \\widehat{KCD}=\\widehat{CIK}$ <br\/> $\\Rightarrow \\Delta CIK$ c\u00e2n t\u1ea1i $K$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A <\/span><\/span>","column":2}]}],"id_ques":1529},{"time":24,"part":[{"time":3,"title":"<span class='basic_left'> Cho \u0111\u01b0\u1eddng tr\u00f2n $(O)$, d\u00e2y cung $AB$. G\u1ecdi $I$ l\u00e0 \u0111i\u1ec3m thu\u1ed9c d\u00e2y $AB$ ($IA > IB$), $D$ l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa c\u1ee7a cung $AB$. Tia $DI$ c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n \u1edf $C$, ti\u1ebfp tuy\u1ebfn c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n t\u1ea1i $C$ c\u1eaft $AB$ t\u1ea1i $K$. <br\/> <b> C\u00e2u 2:<\/b> G\u1ecdi $E$ l\u00e0 \u0111i\u1ec3m \u0111\u1ed1i x\u1ee9ng v\u1edbi $I$ qua $K$. $EC$ c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n $(O)$ \u1edf $M$. Ch\u1ee9ng minh ba \u0111i\u1ec3m $M, O, D$ th\u1eb3ng h\u00e0ng. ","title_trans":"S\u1eafp x\u1ebfp c\u00e1c c\u00e2u \u0111\u1ec3 \u0111\u01b0\u1ee3c b\u00e0i ch\u1ee9ng minh","temp":"sequence","correct":[[[2],[4],[5],[1],[3]]],"list":[{"point":10,"left":[" $\\Rightarrow \\Delta CIE$ vu\u00f4ng t\u1ea1i $C$ (t\u00ednh ch\u1ea5t trung tuy\u1ebfn trong tam gi\u00e1c vu\u00f4ng) ","$\\Rightarrow DM$ l\u00e0 \u0111\u01b0\u1eddng k\u00ednh c\u1ee7a $(O)$ (h\u1ec7 qu\u1ea3 g\u00f3c n\u1ed9i ti\u1ebfp) ","$\\Rightarrow D,\\,O,\\,M$ th\u1eb3ng h\u00e0ng"," X\u00e9t tam gi\u00e1c $CIE$ c\u00f3: $\\left\\{\\begin{align} & KI=KC \\, \\left( \\text{t\u00ednh ch\u1ea5t tam gi\u00e1c c\u00e2n}\\right) \\\\ & KI=KE \\, \\left(\\text{gi\u1ea3 thi\u1ebft}\\right) \\\\ \\end{align}\\right.$ <br\/> $\\Rightarrow KC=KI=KE$"," $\\Rightarrow \\widehat{CIE}={{90}^{o}}$ $\\Rightarrow \\widehat{DCM}={{90}^{o}}$ "],"top":60,"explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai16/lv3/img\/h934_K10.png' \/><\/center> <br\/> X\u00e9t tam gi\u00e1c $CIE$ c\u00f3: $\\left\\{\\begin{align} & KI=KC \\, \\left( \\text{t\u00ednh ch\u1ea5t tam gi\u00e1c c\u00e2n}\\right) \\\\ & KI=KE \\, \\left(\\text{gi\u1ea3 thi\u1ebft}\\right) \\\\ \\end{align}\\right.$ <br\/> $\\Rightarrow KC=KI=KE$ <br\/> $\\Rightarrow \\Delta CIE$ vu\u00f4ng t\u1ea1i $C$ (t\u00ednh ch\u1ea5t trung tuy\u1ebfn trong tam gi\u00e1c vu\u00f4ng) <br\/> $\\Rightarrow \\widehat{ICE}={{90}^{o}}$ $\\Rightarrow \\widehat{DCM}={{90}^{o}}$ <br\/> $\\Rightarrow DM$ l\u00e0 \u0111\u01b0\u1eddng k\u00ednh c\u1ee7a $(O)$ (h\u1ec7 qu\u1ea3 g\u00f3c n\u1ed9i ti\u1ebfp)<br\/> $\\Rightarrow D,\\,O,\\,M$ th\u1eb3ng h\u00e0ng <\/span>"}]}],"id_ques":1530}],"lesson":{"save":0,"level":3}}