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{"segment":[{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","temp":"fill_the_blank","correct":[[["12,5"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"<span class='basic_left'> Cho \u0111\u01b0\u1eddng tr\u00f2n $(O; R)$ v\u00e0 d\u00e2y $AB, \\,C$ l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa cung $AB,\\, M$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a d\u00e2y $AB.$ T\u00ednh b\u00e1n k\u00ednh c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n $(O)$ bi\u1ebft $AB= 24 cm,\\, CM= 9cm.$ <br\/> <b> \u0110\u00e1p s\u1ed1: <\/b> B\u00e1n k\u00ednh \u0111\u01b0\u1eddng tr\u00f2n l\u00e0 _input_ $(cm)$","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai14/lv3/img\/h932_K1.png' \/><\/center> <br\/> K\u1ebb \u0111\u01b0\u1eddng k\u00ednh $CD$ c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n $(O)$ <br\/> Do $\\overset\\frown{AC}=\\overset\\frown{BC}\\Rightarrow AC = BC$ (\u0111\u1ecbnh l\u00ed li\u00ean h\u1ec7 gi\u1eefa d\u00e2y v\u00e0 cung) <br\/> $\\Rightarrow C$ thu\u1ed9c \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $AB$ <br\/> M\u00e0 $MA=MB = 12 cm \\Rightarrow M$ thu\u1ed9c \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $AB$ <br\/> $\\Rightarrow CM$ l\u00e0 trung tr\u1ef1c c\u1ee7a $AB$ <br\/> M\u00e0 $CD$ l\u00e0 trung tr\u1ef1c c\u1ee7a $AB$ (\u0111\u1ecbnh l\u00ed \u0111\u01b0\u1eddng k\u00ednh v\u00e0 d\u00e2y cung) <br\/> $\\Rightarrow C,\\, M,\\, D$ th\u1eb3ng h\u00e0ng <br\/> $\\Rightarrow CD\\bot AB$ <br\/> X\u00e9t $\\Delta CBD$ c\u00f3: <br\/> $OB=OC=OD=R$ <br\/> $\\Rightarrow \\Delta CBD$ vu\u00f4ng t\u1ea1i $B$ (t\u00ednh ch\u1ea5t trung tuy\u1ebfn trong tam gi\u00e1c vu\u00f4ng) <br\/> $\\Rightarrow M{{B}^{2}}=CM.MD$ (h\u1ec7 th\u1ee9c l\u01b0\u1ee3ng trong tam gi\u00e1c vu\u00f4ng) <br\/> $\\Rightarrow MD=\\dfrac{M{{B}^{2}}}{CM} = \\dfrac{{{12}^{2}}}{9}=16\\,\\,\\left( cm \\right)$ <br\/> $\\Rightarrow R=\\dfrac{CD}{2}=\\dfrac{\\left( CM+MD \\right)}{2}=\\dfrac{9+16}{2}=12,5\\,\\,\\left( cm \\right)$ <br\/> <span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $12,5$ <\/span><\/span> "}]}],"id_ques":1461},{"time":24,"part":[{"title":"\u0110i\u1ec1n d\u1ea5u (>; <; =) th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u","temp":"fill_the_blank","correct":[[["<"]]],"list":[{"point":10,"width":50,"type_input":"","ques":" <span class='basic_left'> Cho \u0111\u01b0\u1eddng tr\u00f2n $(O)$ \u0111\u01b0\u1eddng k\u00ednh $AB$ v\u00e0 \u0111\u01b0\u1eddng tr\u00f2n $(O\u2019)$ \u0111\u01b0\u1eddng k\u00ednh $AO.$ L\u1ea5y c\u00e1c \u0111i\u1ec3m $C, D$ thu\u1ed9c \u0111\u01b0\u1eddng tr\u00f2n $(O)$ sao cho $B\\in \\overset\\frown{CD}$ , $\\overset\\frown{BC}<\\overset\\frown{BD}$ . C\u00e1c d\u00e2y $AC,\\, AD$ c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n $(O\u2019)$ theo th\u1ee9 t\u1ef1 $E$ v\u00e0 $F$. <br\/> <b> C\u00e2u 1: <\/b> So s\u00e1nh \u0111\u1ed9 d\u00e0i $OE$ v\u00e0 $OF$<br\/> <b> \u0110\u00e1p s\u1ed1: <\/b> $OE$ _input_ $OF$","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai14/lv3/img\/h932_K2.png' \/><\/center> <br\/> X\u00e9t $\\Delta AEO$ c\u00f3: <br\/> $O'A=O'E=O'O$ (c\u00f9ng b\u1eb1ng b\u00e1n k\u00ednh c\u1ee7a $(O\u2019)$) <br\/> $\\Rightarrow \\Delta AEO$ vu\u00f4ng t\u1ea1i $E$ (t\u00ednh ch\u1ea5t trung tuy\u1ebfn trong tam gi\u00e1c vu\u00f4ng) <br\/> $\\Rightarrow OE\\bot AC$ <br\/> Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1, ta \u0111\u01b0\u1ee3c: $OF \\bot AD $ <br\/> Trong $(O)$ ta c\u00f3: <br\/> $OE\\bot AC \\Rightarrow AE=EC$ (quan h\u1ec7 gi\u1eefa \u0111\u01b0\u1eddng k\u00ednh v\u00e0 d\u00e2y cung) <br\/> $OF\\bot AD \\Rightarrow AF = FD$ (quan h\u1ec7 gi\u1eefa \u0111\u01b0\u1eddng k\u00ednh v\u00e0 d\u00e2y cung) <br\/> $\\Rightarrow OE$ l\u00e0 \u0111\u01b0\u1eddng trung b\u00ecnh c\u1ee7a $\\Delta ABC$ <br\/> $\\Rightarrow OE=\\dfrac{1}{2}BC$ (1) <br\/> T\u01b0\u01a1ng t\u1ef1: $OF=\\dfrac{1}{2}BD$ (2) <br\/> Ta c\u00f3 $\\overset\\frown{BC} < \\overset\\frown{BD} \\Rightarrow BC < BD$ (\u0111\u1ecbnh l\u00fd li\u00ean h\u1ec7 gi\u1eefa cung v\u00e0 d\u00e2y) (3) <br\/> T\u1eeb (1), (2) v\u00e0 (3) $\\Rightarrow OE < OF$ <br\/> <span class='basic_pink'> V\u1eady d\u1ea5u c\u1ea7n \u0111i\u1ec1n l\u00e0 $<$ <\/span><\/span> "}]}],"id_ques":1462},{"time":24,"part":[{"title":"\u0110i\u1ec1n d\u1ea5u (>; <; =) th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u","temp":"fill_the_blank","correct":[[[">"]]],"list":[{"point":10,"width":50,"type_input":"","ques":" <span class='basic_left'> Cho \u0111\u01b0\u1eddng tr\u00f2n $(O)$ \u0111\u01b0\u1eddng k\u00ednh $AB$ v\u00e0 \u0111\u01b0\u1eddng tr\u00f2n $(O\u2019)$ \u0111\u01b0\u1eddng k\u00ednh $AO.$ L\u1ea5y c\u00e1c \u0111i\u1ec3m $C, D$ thu\u1ed9c \u0111\u01b0\u1eddng tr\u00f2n $(O)$ sao cho $B\\in \\overset\\frown{CD}$ , $\\overset\\frown{BC}<\\overset\\frown{BD}$ . C\u00e1c d\u00e2y $AC, \\,AD$ c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n $(O\u2019)$ theo th\u1ee9 t\u1ef1 $E$ v\u00e0 $F$. <br\/> <b> C\u00e2u 2: <\/b> So s\u00e1nh c\u00e1c cung $ AE, \\,AF$ c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n $(O\u2019)$ <br\/> <b> \u0110\u00e1p s\u1ed1: <\/b> $\\overset\\frown{AE}$ _input_ $\\overset\\frown{AF}$","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai14/lv3/img\/h932_K2.png' \/><\/center> <br\/> Trong $(O\u2019)$, ta c\u00f3: <br\/> $\\overset\\frown{AE}+\\overset\\frown{EO}=\\overset\\frown{\\text{OF}}+\\overset\\frown{\\text{AF}}$ (c\u00f9ng b\u1eb1ng n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n) <br\/> M\u00e0 $OE < OF$ (theo c\u00e2u 1) <br\/> $\\Rightarrow \\overset\\frown{EO} < \\overset\\frown{OF}$ (\u0111\u1ecbnh l\u00fd li\u00ean h\u1ec7 gi\u1eefa cung v\u00e0 d\u00e2y) <br\/> Suy ra $\\overset\\frown{AE} > \\overset\\frown{AF}$ <br\/> <span class='basic_pink'> V\u1eady d\u1ea5u c\u1ea7n \u0111i\u1ec1n l\u00e0 $>$ <\/span><\/span> "}]}],"id_ques":1463},{"time":24,"part":[{"time":3,"title":"<span class='basic_left'> Cho \u0111\u01b0\u1eddng tr\u00f2n $(O; R)$ hai d\u00e2y $AB, CD$ vu\u00f4ng g\u00f3c v\u1edbi nhau t\u1ea1i $I\\, (C$ thu\u1ed9c cung nh\u1ecf $AB).$ K\u1ebb \u0111\u01b0\u1eddng k\u00ednh $BE.$ <br\/> <b> C\u00e2u 1: <\/b> Ch\u1ee9ng minh r\u1eb1ng $AC=DE$","title_trans":"S\u1eafp x\u1ebfp c\u00e1c c\u00e2u \u0111\u1ec3 \u0111\u01b0\u1ee3c b\u00e0i ch\u1ee9ng minh","temp":"sequence","correct":[[[2],[3],[5],[1],[4]]],"list":[{"point":10,"left":["$\\Rightarrow \\Delta AEB$ vu\u00f4ng t\u1ea1i $A$ (t\u00ednh ch\u1ea5t trung tuy\u1ebfn trong tam gi\u00e1c vu\u00f4ng) <br\/> $\\Rightarrow AE\\bot AB$ ","M\u00e0 $CD\\bot AB$ $\\Rightarrow AE\/\/CD$ (t\u1eeb vu\u00f4ng g\u00f3c \u0111\u1ebfn song song)"," $\\Rightarrow AC=ED$ (\u0111\u1ecbnh l\u00ed li\u00ean h\u1ec7 gi\u1eefa cung v\u00e0 d\u00e2y) "," X\u00e9t $\\Delta AEB$ c\u00f3: $OA=OE=OB=R$ "," $\\Rightarrow \\overset\\frown{AC}=\\overset\\frown{ED}$ (hai cung b\u1ecb ch\u1eafn gi\u1eefa hai d\u00e2y song song) "],"top":100,"explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai14/lv3/img\/h932_K4.png' \/><\/center> <br\/> X\u00e9t $\\Delta AEB$ c\u00f3: <br\/> $OA=OE=OB=R$ <br\/> $\\Rightarrow \\Delta AEB$ vu\u00f4ng t\u1ea1i $A$ (t\u00ednh ch\u1ea5t trung tuy\u1ebfn trong tam gi\u00e1c vu\u00f4ng) <br\/> $\\Rightarrow AE\\bot AB$ <br\/> M\u00e0 $CD\\bot AB$ $\\Rightarrow AE\/\/CD$ (t\u1eeb vu\u00f4ng g\u00f3c \u0111\u1ebfn song song) <br\/> $\\Rightarrow \\overset\\frown{AC}=\\overset\\frown{ED}$ (hai cung b\u1ecb ch\u1eafn gi\u1eefa hai d\u00e2y song song) <br\/> $\\Rightarrow AC=ED$ (\u0111\u1ecbnh l\u00ed li\u00ean h\u1ec7 gi\u1eefa cung v\u00e0 d\u00e2y) <\/span>"}]}],"id_ques":1464},{"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":[[["4"]]],"list":[{"point":10,"width":50,"type_input":"","input_hint":["frac"],"ques":" <span class='basic_left'> Cho \u0111\u01b0\u1eddng tr\u00f2n $(O; R)$ hai d\u00e2y $AB, CD$ vu\u00f4ng g\u00f3c v\u1edbi nhau t\u1ea1i $I\\, (C$ thu\u1ed9c cung nh\u1ecf $AB).$ K\u1ebb \u0111\u01b0\u1eddng k\u00ednh $BE.$ <br\/> <b> C\u00e2u 2: <\/b> Khi \u0111\u00f3, ta c\u00f3: $I{{A}^{2}}+I{{B}^{2}}+I{{C}^{2}}+I{{D}^{2}}=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}{{R}^{2}}$ ","explain":"<span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai14/lv3/img\/h932_K5.png' \/><\/center> <br\/><br\/> X\u00e9t $\\Delta AIC$ vu\u00f4ng t\u1ea1i $I$ c\u00f3: <br\/> $I{{A}^{2}}+I{{C}^{2}}=A{{C}^{2}}$ (\u0111\u1ecbnh l\u00ed Pitago) <br\/> Ta c\u00f3: $AC =DE$ (theo c\u00e2u 1) <br\/> $\\Rightarrow I{{A}^{2}}+I{{C}^{2}}=D{{E}^{2}}$ (1) <br\/> X\u00e9t $\\Delta BID$ vu\u00f4ng t\u1ea1i $I$ c\u00f3: <br\/> $I{{B}^{2}}+I{{D}^{2}}=B{{D}^{2}}$ (\u0111\u1ecbnh l\u00ed Pitago) (2) <br\/> X\u00e9t $\\Delta BED$ c\u00f3: $OB=OD=OE=R$ <br\/> $\\Rightarrow \\Delta BED$ vu\u00f4ng t\u1ea1i $D$ <br\/> $\\Rightarrow D{{E}^{2}}+D{{B}^{2}}=B{{E}^{2}}={{\\left( 2R \\right)}^{2}}=4{{R}^{2}}$ (\u0111\u1ecbnh l\u00ed Pitago) (3) <br\/> T\u1eeb (1), (2) v\u00e0 (3), suy ra $I{{A}^{2}}+I{{B}^{2}}+I{{C}^{2}}+I{{D}^{2}}=4{{R}^{2}}$ <br\/> <span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $4$<\/span> <\/span>"}]}],"id_ques":1465},{"time":24,"part":[{"time":3,"title":"<span class='basic_left'> Qua \u0111i\u1ec3m $M$ n\u1eb1m ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n $(O),$ v\u1ebd hai c\u00e1t tuy\u1ebfn $MAB$ v\u00e0 $MCD$ v\u1edbi \u0111\u01b0\u1eddng tr\u00f2n sao cho $AB > CD.$ G\u1ecdi $H$ v\u00e0 $K$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AB$ v\u00e0 $CD.$ <br\/> <b> C\u00e2u 1: <\/b> Ch\u1ee9ng minh $MH > MK$ ","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],[6],[1],[4],[2]]],"list":[{"point":10,"left":["X\u00e9t $\\Delta OMH$ vu\u00f4ng t\u1ea1i $H$ c\u00f3: $M{{H}^{2}}+O{{H}^{2}}=O{{M}^{2}}$ (\u0111\u1ecbnh l\u00ed Pitago) (1) ","T\u1eeb (1) v\u00e0 (2) $\\Rightarrow M{{H}^{2}}+O{{H}^{2}}=O{{K}^{2}}+M{{K}^{2}}$ "," Do $OH < OK$ (ch\u1ee9ng minh tr\u00ean) $\\Rightarrow MH > MK$ "," Ta c\u00f3: $AH=HB, CK=CD$ (gi\u1ea3 thi\u1ebft) $\\Rightarrow $ $OH\\bot AB;\\,OK\\bot CD$ (quan h\u1ec7 gi\u1eefa \u0111\u01b0\u1eddng k\u00ednh v\u00e0 d\u00e2y cung) "," X\u00e9t $\\Delta OKM$ vu\u00f4ng t\u1ea1i $K$ c\u00f3: $O{{K}^{2}}+M{{K}^{2}}=O{{M}^{2}}$(\u0111\u1ecbnh l\u00ed Pitago) (2) "," M\u00e0 $AB > CD$ (gi\u1ea3 thi\u1ebft) $\\Rightarrow OH < OK$ (li\u00ean h\u1ec7 gi\u1eefa d\u00e2y cung v\u00e0 kho\u1ea3ng c\u00e1ch \u0111\u1ebfn t\u00e2m) "],"top":100,"explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai14/lv3/img\/h932_K6.png' \/><\/center> <br\/> Ta c\u00f3: $AH=HB, CK=CD$ (gi\u1ea3 thi\u1ebft) <br\/> $\\Rightarrow $ $OH\\bot AB;\\,OK\\bot CD$ (quan h\u1ec7 gi\u1eefa \u0111\u01b0\u1eddng k\u00ednh v\u00e0 d\u00e2y cung) <br\/> M\u00e0 $AB > CD$ (gi\u1ea3 thi\u1ebft) <br\/> $\\Rightarrow OH < OK$ (li\u00ean h\u1ec7 gi\u1eefa d\u00e2y cung v\u00e0 kho\u1ea3ng c\u00e1ch \u0111\u1ebfn t\u00e2m) <br\/> X\u00e9t $\\Delta OMH$ vu\u00f4ng t\u1ea1i $H$ c\u00f3: <br\/> $M{{H}^{2}}+O{{H}^{2}}=O{{M}^{2}}$ (\u0111\u1ecbnh l\u00ed Pitago) (1) <br\/> X\u00e9t $\\Delta OKM$ vu\u00f4ng t\u1ea1i $K$ c\u00f3: <br\/> $O{{K}^{2}}+M{{K}^{2}}=O{{M}^{2}}$(\u0111\u1ecbnh l\u00ed Pitago) (2) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow M{{H}^{2}}+O{{H}^{2}}=O{{K}^{2}}+M{{K}^{2}}$ <br\/> Do $OH < OK$ (ch\u1ee9ng minh tr\u00ean) $\\Rightarrow MH > MK$ <\/span>"}]}],"id_ques":1466},{"time":24,"part":[{"time":3,"title":"<span class='basic_left'> Qua \u0111i\u1ec3m $M$ n\u1eb1m ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n $(O),$ v\u1ebd hai c\u00e1t tuy\u1ebfn $MAB$ v\u00e0 $MCD$ v\u1edbi \u0111\u01b0\u1eddng tr\u00f2n sao cho $AB > CD.$ G\u1ecdi $H$ v\u00e0 $K$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AB$ v\u00e0 $CD.$ <br\/> <b> C\u00e2u 2: <\/b> Ch\u1ee9ng minh $\\widehat{MOH}>\\widehat{MOK}$","title_trans":"S\u1eafp x\u1ebfp c\u00e1c c\u00e2u \u0111\u1ec3 \u0111\u01b0\u1ee3c b\u00e0i ch\u1ee9ng minh ","temp":"sequence","correct":[[[4],[1],[5],[3],[6],[2]]],"list":[{"point":10,"left":[" $\\Rightarrow IM=IO=IH=IK$ <br\/> $\\Rightarrow H,M,K,O$ c\u00f9ng thu\u1ed9c \u0111\u01b0\u1eddng tr\u00f2n $(I)$ ","G\u1ecdi $I$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $MO$. V\u1ebd \u0111\u01b0\u1eddng tr\u00f2n t\u00e2m $(I; IO)$ "," Trong \u0111\u01b0\u1eddng tr\u00f2n $(I)$, ta c\u00f3: $MH > MK$ (theo c\u00e2u 1) <br\/> $\\Rightarrow \\overset\\frown{MH}>\\overset\\frown{MK}$ (\u0111\u1ecbnh l\u00ed li\u00ean h\u1ec7 gi\u1eefa cung v\u00e0 d\u00e2y) "," Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1 $\\Rightarrow IK=IM=IO$ "," Hay $\\text{s\u0111}\\overset\\frown{MH}>\\text{s\u0111}\\overset\\frown{MK}$ $\\Rightarrow \\widehat{MIH}=\\dfrac{1}{2}\\text{s\u0111}\\overset\\frown{MH}>\\dfrac{1}{2}\\text{s\u0111}\\overset\\frown{MH}=\\widehat{MIK}$ (\u0111\u1ecbnh l\u00ed g\u00f3c n\u1ed9i ti\u1ebfp) ","X\u00e9t $\\Delta MHO$ vu\u00f4ng t\u1ea1i $H$ c\u00f3: $IO=IM$ (c\u00e1ch d\u1ef1ng) <br\/> $\\Rightarrow IO=IM=IH$ (t\u00ednh ch\u1ea5t trung tuy\u1ebfn trong tam gi\u00e1c vu\u00f4ng)"],"top":100,"hint":"G\u1ecdi $I$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $MO$. V\u1ebd \u0111\u01b0\u1eddng tr\u00f2n t\u00e2m $(I; IO)$ r\u1ed3i ch\u1ee9ng minh $\\text{s\u0111}\\overset\\frown{MH}>\\text{s\u0111}\\overset\\frown{MK}$","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai14/lv3/img\/h932_K7.png' \/><\/center> <br\/> G\u1ecdi $I$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $MO$. V\u1ebd \u0111\u01b0\u1eddng tr\u00f2n t\u00e2m $(I; IO)$ <br\/> X\u00e9t $\\Delta MHO$ vu\u00f4ng t\u1ea1i $H$ c\u00f3: <br\/> $IO=IM$ (c\u00e1ch d\u1ef1ng) <br\/> $\\Rightarrow IO=IM=IH$ (t\u00ednh ch\u1ea5t trung tuy\u1ebfn trong tam gi\u00e1c vu\u00f4ng) <br\/> Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1 $\\Rightarrow IK=IM=IO$ <br\/> $\\Rightarrow IM=IO=IH=IK$ <br\/> $\\Rightarrow H,M,K,O$ c\u00f9ng thu\u1ed9c \u0111\u01b0\u1eddng tr\u00f2n $(I)$ <br\/> Trong \u0111\u01b0\u1eddng tr\u00f2n $(I)$, ta c\u00f3: <br\/> $MH > MK$ (theo c\u00e2u 1) <br\/> $\\Rightarrow \\overset\\frown{MH}>\\overset\\frown{MK}$ (\u0111\u1ecbnh l\u00ed li\u00ean h\u1ec7 gi\u1eefa cung v\u00e0 d\u00e2y) <br\/> Hay $\\text{s\u0111}\\overset\\frown{MH}>\\text{s\u0111}\\overset\\frown{MK}$ <br\/> $\\Rightarrow \\widehat{MIH}=\\dfrac{1}{2}\\text{s\u0111}\\overset\\frown{MH}>\\dfrac{1}{2}\\text{s\u0111}\\overset\\frown{MH}=\\widehat{MIK}$ (\u0111\u1ecbnh l\u00ed g\u00f3c n\u1ed9i ti\u1ebfp) <\/span>"}]}],"id_ques":1467},{"time":24,"part":[{"time":3,"title":"<span class='basic_left'> Cho h\u00ecnh v\u1ebd sau, <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai14/lv3/img\/h932_K8.png' \/><\/center> <br\/> Trong \u0111\u00f3, $MN = PQ$. Ch\u1ee9ng minh r\u1eb1ng $AN = AQ$","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],[2],[6],[4]]],"list":[{"point":10,"left":[" $\\Rightarrow \\Delta OEA=\\Delta OFA$ (c\u1ea1nh huy\u1ec1n v\u00e0 c\u1ea1nh g\u00f3c nh\u1ecdn) <br\/> $\\Rightarrow AE=AF$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) ","M\u00e0 $MN = PQ$ (gi\u1ea3 thi\u1ebft) $\\Rightarrow EN=FQ$ "," Ta c\u00f3: $MN=PQ\\Rightarrow OE=OF$ (\u0111\u1ecbnh l\u00ed kho\u1ea3ng c\u00e1ch t\u1eeb t\u00e2m \u0111\u1ebfn d\u00e2y) "," X\u00e9t hai tam gi\u00e1c vu\u00f4ng $\\Delta OEA$ v\u00e0 $\\Delta OFA$ c\u00f3: $\\left\\{ \\begin{align} & OE=OF\\,\\left( \\text{ch\u1ee9ng minh tr\u00ean} \\right) \\\\ & OA\\,\\text{chung} \\\\ \\end{align} \\right.$ "," $\\Rightarrow AE-EN=AQ-FQ\\,hay\\,AN=AQ$ ","V\u00ec $\\left\\{ \\begin{align} & OE\\bot MN \\\\ & OF\\bot PQ \\\\ \\end{align} \\right.$$\\Rightarrow \\left\\{ \\begin{align} & ME=EN \\\\ & PF=FQ \\\\ \\end{align} \\right.$ (\u0111\u1ecbnh l\u00ed \u0111\u01b0\u1eddng k\u00ednh v\u00e0 d\u00e2y cung) "],"top":100,"explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai14/lv3/img\/h932_K8.png' \/><\/center> <br\/> Ta c\u00f3: $MN=PQ$ <br\/> $\\Rightarrow OE=OF$ (\u0111\u1ecbnh l\u00ed kho\u1ea3ng c\u00e1ch t\u1eeb t\u00e2m \u0111\u1ebfn d\u00e2y) <br\/> X\u00e9t hai tam gi\u00e1c vu\u00f4ng $\\Delta OEA$ v\u00e0 $\\Delta OFA$ c\u00f3: <br\/> $\\left\\{ \\begin{align} & OE=OF\\,\\left(\\text{ch\u1ee9ng minh tr\u00ean} \\right) \\\\ & OA\\,\\text{chung} \\\\ \\end{align} \\right.$ <br\/> $\\Rightarrow \\Delta OEA=\\Delta OFA$ (c\u1ea1nh huy\u1ec1n v\u00e0 c\u1ea1nh g\u00f3c nh\u1ecdn) <br\/> $\\Rightarrow AE=AF$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) <br\/> V\u00ec $\\left\\{ \\begin{align} & OE\\bot MN \\\\ & OF\\bot PQ \\\\ \\end{align} \\right.$$\\Rightarrow \\left\\{ \\begin{align} & ME=EN \\\\ & PF=FQ \\\\ \\end{align} \\right.$ (\u0111\u1ecbnh l\u00ed \u0111\u01b0\u1eddng k\u00ednh v\u00e0 d\u00e2y cung) <br\/> M\u00e0 $MN = PQ$ (gi\u1ea3 thi\u1ebft) <br\/> $\\Rightarrow EN=FQ$ <br\/> $\\Rightarrow AE-EN=AF-FQ$ hay $AN=AQ$ <\/span>"}]}],"id_ques":1468},{"time":24,"part":[{"time":3,"title":"Cho \u0111\u01b0\u1eddng tr\u00f2n $(O)$, hai d\u00e2y $AB, CD$ b\u1eb1ng nhau v\u00e0 c\u1eaft nhau t\u1ea1i \u0111i\u1ec3m $I$ n\u1eb1m trong \u0111\u01b0\u1eddng tr\u00f2n. Ch\u1ee9ng minh r\u1eb1ng $OI$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a m\u1ed9t trong hai g\u00f3c t\u1ea1o b\u1edfi hai d\u00e2y $AB$ v\u00e0 $CD$ ","title_trans":"S\u1eafp x\u1ebfp c\u00e1c c\u00e2u \u0111\u1ec3 \u0111\u01b0\u1ee3c b\u00e0i ch\u1ee9ng minh","temp":"sequence","correct":[[[2],[4],[5],[6],[3],[1]]],"list":[{"point":10,"left":[" V\u00ec $AB=CD\\Rightarrow OH=OK$ (\u0111\u1ecbnh l\u00ed kho\u1ea3ng c\u00e1ch t\u1eeb t\u00e2m \u0111\u1ebfn d\u00e2y) ","$\\left\\{ \\begin{align} & OH=OK\\left( \\text{ch\u1ee9ng minh tr\u00ean} \\right) \\\\ & OI\\,\\text{chung} \\\\ \\end{align} \\right.$ "," $\\Rightarrow \\Delta OHI=\\Delta OKI$ (c\u1ea1nh huy\u1ec1n v\u00e0 c\u1ea1nh g\u00f3c vu\u00f4ng) <br\/> $\\Rightarrow \\widehat{OIH}=\\widehat{OIK}$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) "," $\\Rightarrow OI$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a m\u1ed9t g\u00f3c t\u1ea1o th\u00e0nh b\u1edfi hai d\u00e2y $AB$ v\u00e0 $CD$ "," X\u00e9t hai tam gi\u00e1c vu\u00f4ng $\\Delta OHI$ v\u00e0 $\\Delta OKI$ c\u00f3: ","K\u1ebb c\u00e1c \u0111\u01b0\u1eddng th\u1eb3ng $OH\\bot AB;\\,OK\\bot CD$ "],"top":100,"explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai14/lv3/img\/h932_K9.png' \/><\/center> <br\/> K\u1ebb c\u00e1c \u0111\u01b0\u1eddng th\u1eb3ng $OH\\bot AB;\\,OK\\bot CD$ <br\/> V\u00ec $AB=CD \\Rightarrow OH=OK$ (\u0111\u1ecbnh l\u00ed kho\u1ea3ng c\u00e1ch t\u1eeb t\u00e2m \u0111\u1ebfn d\u00e2y) <br\/> X\u00e9t hai tam gi\u00e1c vu\u00f4ng $\\Delta OHI$ v\u00e0 $\\Delta OKI$ c\u00f3: <br\/> $\\left\\{ \\begin{align} & OH=OK\\left( \\text{ch\u1ee9ng minh tr\u00ean} \\right) \\\\ & OI\\,\\text{chung} \\\\ \\end{align} \\right.$ <br\/> $\\Rightarrow \\Delta OHI=\\Delta OKI$ (c\u1ea1nh huy\u1ec1n v\u00e0 c\u1ea1nh g\u00f3c vu\u00f4ng) <br\/> $\\Rightarrow \\widehat{OIH}=\\widehat{OIK}$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) <br\/> $\\Rightarrow OI$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a m\u1ed9t g\u00f3c t\u1ea1o th\u00e0nh b\u1edfi hai d\u00e2y $AB$ v\u00e0 $CD$ <\/span>"}]}],"id_ques":1469},{"time":24,"part":[{"time":3,"title":"Cho \u0111\u01b0\u1eddng tr\u00f2n $(O;R)$ v\u00e0 c\u00e1c b\u00e1n k\u00ednh $OA; OB$. Tr\u00ean cung nh\u1ecf $AB$ l\u1ea5y c\u00e1c \u0111i\u1ec3m $M$ v\u00e0 $N$ sao cho $AM = BN$. G\u1ecdi $C$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $BN$ v\u00e0 $AM$, $OC$ c\u1eaft $(O)$ t\u1ea1i $E.$ Ch\u1ee9ng minh $\\overset\\frown{AE}=\\overset\\frown{BE}$. ","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],[6],[4],[2]]],"list":[{"point":10,"left":[" $\\Rightarrow \\Delta OAH=\\Delta OBK$ (c\u1ea1nh huy\u1ec1n v\u00e0 c\u1ea1nh g\u00f3c vu\u00f4ng) <br\/> $\\Rightarrow \\widehat{{{O}_{1}}}=\\widehat{{{O}_{4}}}$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng)","$\\Rightarrow \\widehat{{{O}_{1}}}+\\widehat{{{O}_{2}}}=\\widehat{{{O}_{3}}}+\\widehat{{{O}_{4}}}\\Rightarrow \\widehat{AOC}=\\widehat{BOC}$"," K\u1ebb $OH\\bot AM;OK\\bot BN$ <br\/> V\u00ec $AM=BN\\Rightarrow OH=OK$ (\u0111\u1ecbnh l\u00ed kho\u1ea3ng c\u00e1ch t\u1eeb t\u00e2m \u0111\u1ebfn d\u00e2y)"," $\\Rightarrow \\text{s\u0111}\\overset\\frown{AE}=\\text{s\u0111}\\overset\\frown{BE}$ hay $ \\overset\\frown{AE}=\\overset\\frown{BE}$ (\u0111\u1ecbnh l\u00ed li\u00ean h\u1ec7 gi\u1eefa d\u00e2y v\u00e0 cung) "," Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1, ta \u0111\u01b0\u1ee3c: $\\Delta OHC=\\Delta OKC$ <br\/> $\\Rightarrow \\widehat{{{O}_{2}}}=\\widehat{{{O}_{3}}}$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng)"," X\u00e9t hai tam gi\u00e1c vu\u00f4ng $\\Delta OAH$ v\u00e0 $\\Delta OBK$ c\u00f3: $\\left\\{ \\begin{align} & OA=OB=R \\\\ & OH=OK\\,\\left( \\text{ch\u1ee9ng minh tr\u00ean} \\right) \\\\ \\end{align} \\right.$ "],"top":100,"explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai14/lv3/img\/h932_K10.png' \/><\/center> <br\/> K\u1ebb $OH\\bot AM;OK\\bot BN$ <br\/> V\u00ec $AM=BN\\Rightarrow OH=OK$ (\u0111\u1ecbnh l\u00ed kho\u1ea3ng c\u00e1ch t\u1eeb t\u00e2m \u0111\u1ebfn d\u00e2y) <br\/> X\u00e9t hai tam gi\u00e1c vu\u00f4ng $\\Delta OAH$ v\u00e0 $\\Delta OBK$ c\u00f3: <br\/> $\\left\\{ \\begin{align} & OA=OB=R \\\\ & OH=OK\\,\\left( \\text{ch\u1ee9ng minh tr\u00ean} \\right) \\\\ \\end{align} \\right.$ <br\/> $\\Rightarrow \\Delta OAH=\\Delta OBK$ (c\u1ea1nh huy\u1ec1n v\u00e0 c\u1ea1nh g\u00f3c vu\u00f4ng) <br\/> $\\Rightarrow \\widehat{{{O}_{1}}}=\\widehat{{{O}_{4}}}$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) <br\/> Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1, ta \u0111\u01b0\u1ee3c: $\\Delta OHC=\\Delta OKC$ <br\/> $\\Rightarrow \\widehat{{{O}_{2}}}=\\widehat{{{O}_{3}}}$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) <br\/> $\\Rightarrow \\widehat{{{O}_{1}}}+\\widehat{{{O}_{2}}}=\\widehat{{{O}_{3}}}+\\widehat{{{O}_{4}}}$ <br\/> $\\Rightarrow \\widehat{AOC}=\\widehat{BOC}$ <br\/> $\\Rightarrow \\text{s\u0111}\\overset\\frown{AE}=\\text{s\u0111}\\overset\\frown{BE}$ hay $\\overset\\frown{AE}=\\overset\\frown{BE}$ (\u0111\u1ecbnh l\u00ed li\u00ean h\u1ec7 gi\u1eefa d\u00e2y v\u00e0 cung) <\/span>"}]}],"id_ques":1470}],"lesson":{"save":0,"level":3}}

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Câu hỏi này theo dạng chọn đáp án đúng, sau khi đọc xong câu hỏi, bạn bấm vào một trong số các đáp án mà chương trình đưa ra bên dưới, sau đó bấm vào nút gửi để kiểm tra đáp án và sẵn sàng chuyển sang câu hỏi kế tiếp

Trả lời đúng trong khoảng thời gian quy định bạn sẽ được + số điểm như sau:
Trong khoảng 1 phút đầu tiên + 5 điểm
Trong khoảng 1 phút -> 2 phút + 4 điểm
Trong khoảng 2 phút -> 3 phút + 3 điểm
Trong khoảng 3 phút -> 4 phút + 2 điểm
Trong khoảng 4 phút -> 5 phút + 1 điểm

Quá 5 phút: không được cộng điểm

Tổng thời gian làm mỗi câu (không giới hạn)

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