{"segment":[{"time":24,"part":[{"title":"\u0110i\u1ec1n t\u1eeb th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["75"],["15"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","hint":"T\u00ednh $\\widehat{ADH} \\to \\widehat{BAD} \\to \\widehat{B} \\to \\widehat{C}$","ques":" <span class='basic_left'> Cho tam gi\u00e1c $ABC$ c\u00f3 $\\widehat{B} > \\widehat{C} $. K\u1ebb $AH$ vu\u00f4ng g\u00f3c v\u1edbi $BC$ t\u1ea1i H, k\u1ebb \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c $AD$ c\u1ee7a g\u00f3c $\\widehat{A} $ ($D\\in BC$). Bi\u1ebft $\\widehat{HAD} = 30^{o} $ v\u00e0 $\\widehat{A} = 90^{o} $ <br\/> T\u00ednh s\u1ed1 \u0111o g\u00f3c $B$ v\u00e0 g\u00f3c $C$. <br\/> <b> \u0110\u00e1p \u00e1n: <\/b> <br\/> $\\widehat{B} = \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}^{o} \\\\ \\widehat{C} = \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}^{o} $ <\/span> ","explain":"<span class='basic_left'> <span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> T\u00ednh $\\widehat{ADH}$ trong $\\Delta ADH$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> T\u00ednh $\\widehat{BAD}$ <br\/> <b> B\u01b0\u1edbc 3: <\/b> T\u00ednh $\\widehat{B}$ trong $\\Delta ABD$ <br\/> <b> B\u01b0\u1edbc 4: <\/b> T\u00ednh $\\widehat{C}$ <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span> <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai16/lv3/img\/H7B16-26.png' \/> <\/center> * Tam gi\u00e1c $AHD$ vu\u00f4ng t\u1ea1i H n\u00ean $\\widehat{HAD} + \\widehat{ADH} = 90^{o} $ <br\/> $\\Rightarrow \\widehat{ADH} = 90^{o} - \\widehat{HAD} = 90^{o} - 30^{o} = 60^{o} $ hay $\\widehat{ADB} = 60^{o} $ <br\/> * AD l\u00e0 ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c A n\u00ean $\\widehat{BAD} = \\dfrac{\\widehat{A}}{2} = \\dfrac{90^{o}}{2} = 45^{o} $ <br\/> *Tam gi\u00e1c ABD c\u00f3 $\\widehat{BAD} + \\widehat{B} + \\widehat{ADB} = 180^{o} \\\\ \\Leftrightarrow 45^{o} + \\widehat{B} + 60^{o} = 180^{o} \\\\ \\Leftrightarrow \\widehat{B} = 75^{o} $ <br\/> *Tam gi\u00e1c ABC vu\u00f4ng t\u1ea1i A n\u00ean $\\widehat{B} + \\widehat{C} = 90^{o} $ <br\/> m\u00e0 $\\widehat{B} = 75^{o} $ n\u00ean $\\widehat{C} = 90^{o} - 75^{o} = 15^{o}$ <br\/> <br\/> <span class='basic_pink'> \u0110\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u1ea7n l\u01b0\u1ee3t l\u00e0: 75 v\u00e0 15<\/span><\/span> <br\/>"}]}],"id_ques":1781},{"time":24,"part":[{"title":"\u0110i\u1ec1n t\u1eeb th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["80"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","ques":" <span class='basic_left'>Cho tam gi\u00e1c $ABC$, bi\u1ebft $\\widehat{A} = 3\\widehat{B} $ v\u00e0 $\\widehat{B} = 2\\widehat{C} $ <br\/> G\u1ecdi E l\u00e0 giao \u0111i\u1ec3m c\u1ee7a \u0111\u01b0\u1eddng th\u1eb3ng $AB$ v\u1edbi ph\u00e2n gi\u00e1c g\u00f3c ngo\u00e0i \u1edf \u0111\u1ec9nh C c\u1ee7a tam gi\u00e1c $ABC$. <br\/> S\u1ed1 \u0111o g\u00f3c $\\widehat{ACE} $ l\u00e0: $\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}^o$ <\/span> ","explain":"<span class='basic_left'> <span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> T\u00ednh $\\widehat{A}$ v\u00e0 $\\widehat{B}$ trong $\\Delta ABC$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> T\u00ednh $\\widehat{CAx}$ l\u00e0 g\u00f3c ngo\u00e0i $\\Delta ABC$ <br\/> <b> B\u01b0\u1edbc 3: <\/b> T\u00ednh $\\widehat{ACE}$ theo t\u00ednh ch\u1ea5t ph\u00e2n gi\u00e1c <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span> <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai16/lv3/img\/H7B16-27.png' \/> <\/center> <span class='basic_left'> Tam gi\u00e1c $ABC$ c\u00f3: $\\widehat{A} = 3\\widehat{B} $ v\u00e0 $\\widehat{B} = 2\\widehat{C} $ n\u00ean $\\widehat{A} = 3.2\\widehat{C} = 6\\widehat{C} $<br\/> Do $\\widehat{A} + \\widehat{B} + \\widehat{C} = 180^{o} \\\\ \\Rightarrow 6\\widehat{C} + 2\\widehat{C} + \\widehat{C} = 180^{o} \\\\ \\Leftrightarrow 9\\widehat{C} = 180^{o} \\\\ \\Rightarrow \\widehat{C} = 20^{o} \\\\ \\Rightarrow \\widehat{A} = 6.20^{o} = 120^{o}; \\widehat{B} = 2.20^{o} = 40^{o} $ <br\/> Tam gi\u00e1c $ABC$ c\u00f3: $\\widehat{ACx} = \\widehat{A} + \\widehat{B} $ (g\u00f3c ngo\u00e0i c\u1ee7a tam gi\u00e1c) <br\/> $\\Rightarrow \\widehat{ACx} = 120^{o} + 40^{o} = 160^{o} $ <br\/> CE l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{ACx} $ n\u00ean: $\\widehat{ACE} = \\dfrac{1}{2}\\widehat{ACx} = \\dfrac{1}{2}.160^{o} = 80^{o} $ <br\/> <span class='basic_pink'> \u0110\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0: 80<\/span><\/span> <br\/>"}]}],"id_ques":1782},{"time":24,"part":[{"title":"\u0110i\u1ec1n t\u1eeb th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["11"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","ques":" <span class='basic_left'> Cho $\\triangle ABC = \\triangle MNP $. <br\/> Bi\u1ebft $AB + BC = 7cm, MN - NP = 3cm, MP = 4cm.$ <br\/> Chu vi tam gi\u00e1c $ABC$ l\u00e0: $\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}(cm)$ ","explain":"<span class='basic_left'> <span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> T\u00ednh \u0111\u1ed9 d\u00e0i $AC$ theo t\u00ednh ch\u1ea5t hai tam gi\u00e1c $ABC$ v\u00e0 $MNP$ b\u1eb1ng nhau <br\/> <b> B\u01b0\u1edbc 2: <\/b> T\u00ednh $AB$ v\u00e0 $BC$ khi bi\u1ebft t\u1ed5ng v\u00e0 hi\u1ec7u <br\/> <b> B\u01b0\u1edbc 3: <\/b> T\u00ednh chu vi $\\Delta ABC$ <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span> <br\/> Ta c\u00f3: $\\triangle ABC = \\triangle MNP $ <br\/> $\\Rightarrow$ $AB = MN, BC = NP, AC = MP$ (c\u00e1c c\u1eb7p c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) <br\/> M\u00e0 $MN - NP = 3cm$ $\\Rightarrow$ $ AB - BC = 3cm$ <br\/> Theo b\u00e0i c\u00f3 $AB + BC = 7cm$ <br\/> Suy ra: $2AB = 10cm$ $\\Rightarrow $ $AB = 5cm$ <br\/> $\\Rightarrow $ $BC = 7 - AC = 2cm$ <br\/> Do \u0111\u00f3 ta c\u00f3: $AB = 5cm, BC = 2cm, AC = 4cm$ <br\/> V\u1eady chu vi tam gi\u00e1c $ABC$ b\u1eb1ng: <br\/> $AB + BC + AC = 5 + 2 + 4 = 11cm$ <br\/><br\/><span class='basic_pink'>\u0110\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0: 11 <\/span> "}]}],"id_ques":1783},{"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":" <span class='basic_left'> Cho $\\triangle MNP $ c\u00f3 $MN = MP = NP$ v\u00e0 \u0111i\u1ec3m O n\u1eb1m trong tam gi\u00e1c sao cho <br\/> $OM = ON = OP$. S\u1ed1 \u0111o g\u00f3c $\\widehat{NOP} $ l\u00e0: <\/span>","select":[" A. $60^{o} $ "," B. $90^{o} $ "," C. $120^{o} $ ","D. $150^{o} $ "],"explain":"<span class='basic_left'> <span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> Ch\u1ee9ng minh $\\triangle MON = \\triangle PON$ (c.c.c) <br\/> <b> B\u01b0\u1edbc 2: <\/b> Ch\u1ee9ng minh $\\triangle MON = \\triangle MOP$ (c.c.c) <br\/> <b> B\u01b0\u1edbc 3: <\/b> Suy ra: $\\triangle MON = \\triangle PON=\\triangle MOP$ <br\/> <b> B\u01b0\u1edbc 4: <\/b> Ch\u1ec9 ra $\\widehat{MON} = \\widehat{NOP} = \\widehat{MOP} $ v\u00e0 suy ra s\u1ed1 \u0111o g\u00f3c $NOP$ <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span> <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai16/lv3/img\/H7B16-28.png' \/> <\/center> <span class='basic_left'> X\u00e9t $\\triangle MON $ v\u00e0 $\\triangle PON $ c\u00f3: <br\/> + c\u1ea1nh $ON$ chung<br\/> + $ MO = PO$ (gi\u1ea3 thi\u1ebft)<br\/> + $ MN = PN$ (gi\u1ea3 thi\u1ebft) <br\/> $\\Rightarrow \\triangle MON = \\triangle PON$ (c.c.c) (1) <br\/> T\u01b0\u01a1ng t\u1ef1, x\u00e9t $\\triangle MON $ v\u00e0 $\\triangle MOP $ c\u00f3: <br\/> + c\u1ea1nh $MO$ chung <br\/> + $ MN = MP$ (gi\u1ea3 thi\u1ebft)<br\/> + $ON = OP$ (gi\u1ea3 thi\u1ebft) <br\/> $\\Rightarrow \\triangle MON = \\triangle MOP$ (c.c.c) (2) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow \\triangle MON = \\triangle PON = \\triangle MOP $ <br\/> Suy ra $\\widehat{MON} = \\widehat{NOP} = \\widehat{MOP} $ <br\/> M\u00e0 $\\widehat{MON} + \\widehat{NOP} + \\widehat{MOP} = 360^{o} $ <br\/> Do \u0111\u00f3: $3\\widehat{NOP} = 360^{o} \\Rightarrow \\widehat{NOP} = 120^{o} $ <br\/><span class='basic_pink'>\u0110\u00e1p \u00e1n \u0111\u00fang l\u00e0 C. <\/span>","column":4}]}],"id_ques":1784},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00fang hay sai","title_trans":"","temp":"true_false","correct":[["t","t","t","t","f"]],"list":[{"point":10,"image":"","ques":"","col_name":["","\u0110\u00fang","Sai"],"arr_ques":["N\u1ebfu hai tam gi\u00e1c b\u1eb1ng nhau th\u00ec hai c\u1ea1nh v\u00e0 g\u00f3c xen gi\u1eefa c\u1ee7a tam gi\u00e1c n\u00e0y b\u1eb1ng hai c\u1ea1nh v\u00e0 g\u00f3c xen gi\u1eefa c\u1ee7a tam gi\u00e1c kia.","N\u1ebfu hai c\u1ea1nh v\u00e0 g\u00f3c xen gi\u1eefa c\u1ee7a tam gi\u00e1c n\u00e0y b\u1eb1ng hai c\u1ea1nh v\u00e0 g\u00f3c xen gi\u1eefa c\u1ee7a tam gi\u00e1c kia th\u00ec hai tam gi\u00e1c \u0111\u00f3 b\u1eb1ng nhau.","N\u1ebfu hai c\u1ea1nh v\u00e0 g\u00f3c xen gi\u1eefa c\u1ee7a tam gi\u00e1c n\u00e0y b\u1eb1ng hai c\u1ea1nh v\u00e0 g\u00f3c xen gi\u1eefa c\u1ee7a tam gi\u00e1c kia th\u00ec hai c\u1ea1nh c\u00f2n l\u1ea1i c\u0169ng b\u1eb1ng nhau.","N\u1ebfu hai c\u1ea1nh v\u00e0 g\u00f3c xen gi\u1eefa c\u1ee7a tam gi\u00e1c n\u00e0y b\u1eb1ng hai c\u1ea1nh v\u00e0 g\u00f3c xen gi\u1eefa c\u1ee7a tam gi\u00e1c kia th\u00ec hai c\u1eb7p g\u00f3c c\u00f2n l\u1ea1i c\u0169ng t\u01b0\u01a1ng \u1ee9ng b\u1eb1ng nhau.","N\u1ebfu hai tam gi\u00e1c c\u00f3 hai c\u1eb7p g\u00f3c t\u01b0\u01a1ng \u1ee9ng b\u1eb1ng nhau th\u00ec hai tam gi\u00e1c c\u0169ng c\u00f3 hai c\u1eb7p c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng b\u1eb1ng nhau. "],"explain":["<span class='basic_left'>1-\u0110\u00fang. V\u00ec theo \u0111\u1ecbnh ngh\u0129a hai tam gi\u00e1c b\u1eb1ng nhau. <\/span>","<span class='basic_left'>2- \u0110\u00fang v\u00ec theo tr\u01b0\u1eddng h\u1ee3p b\u1eb1ng nhau c\u1ea1nh - g\u00f3c - c\u1ea1nh c\u1ee7a hai tam gi\u00e1c<\/span>","<span class='basic_left'>3- \u0110\u00fang v\u00ec theo tr\u01b0\u1eddng h\u1ee3p b\u1eb1ng nhau c\u1ea1nh - g\u00f3c - c\u1ea1nh c\u1ee7a hai tam gi\u00e1c, ta c\u00f3 hai tam gi\u00e1c b\u1eb1ng nhau n\u00ean hai c\u1ea1nh c\u00f2n l\u1ea1i b\u1eb1ng nhau<\/span>","<span class='basic_left'>4- \u0110\u00fang v\u00ec theo tr\u01b0\u1eddng h\u1ee3p b\u1eb1ng nhau c\u1ea1nh - g\u00f3c - c\u1ea1nh c\u1ee7a hai tam gi\u00e1c, ta c\u00f3 hai tam gi\u00e1c b\u1eb1ng nhau n\u00ean hai c\u1eb7p g\u00f3c t\u01b0\u01a1ng \u1ee9ng c\u00f2n l\u1ea1i b\u1eb1ng nhau<\/span>","<span class='basic_left'>5- Sai, v\u00ed d\u1ee5 hai tam gi\u00e1c $ABC$ v\u00e0 $A'B'C'$ c\u00f3 $\\widehat{B} = \\widehat{B'}, \\widehat{B'} = \\widehat{C} $ nh\u01b0ng kh\u00f4ng c\u00f3 hai c\u1eb7p c\u1ea1nh n\u00e0o b\u1eb1ng nhau. <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai16/lv3/img\/H7B16-29.png' \/><\/center> "]}]}],"id_ques":1785},{"time":24,"part":[{"title":"L\u1ef1a 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 $ABC$ vu\u00f4ng t\u1ea1i $A, M$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $BC$. Ta c\u00f3: ","select":[" A. $AC = \\dfrac{1}{2} BC $"," B. $AM > \\dfrac{1}{2} BC $"," C. $AM < \\dfrac{1}{2} BC $ ","D. $AM = \\dfrac{1}{2} BC $ "],"explain":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai16/lv3/img\/H7B16-30.png' \/><\/center> <span class='basic_left'>Tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia $MA$ l\u1ea5y D sao cho $MD = MA = \\dfrac{DA}{2} $ <br\/> X\u00e9t $\\triangle AMB $ v\u00e0 $\\triangle DMC $ c\u00f3: <br\/> + $MA = MD$ (theo c\u00e1ch v\u1ebd \u0111i\u1ec3m D) <br\/> + $\\widehat{AMB} = \\widehat{DMC} $ (hai g\u00f3c \u0111\u1ed1i \u0111\u1ec9nh) <br\/> + $ MB = MC$ (v\u00ec M l\u00e0 trung \u0111i\u1ec3m c\u1ee7a BC) <br\/> Do \u0111\u00f3 $\\triangle AMB = \\triangle DMC $ (c - g - c) <br\/> $\\Rightarrow AB = DC, \\widehat{MAB} = \\widehat{MDC} $ <br\/> M\u00e0 $\\widehat{MAB}$ v\u00e0 $\\widehat{MDC}$ so le trong n\u00ean $AB \/\/ CD$ <br\/> V\u00ec $AB \\perp AC $ ($\\triangle ABC$ vu\u00f4ng t\u1ea1i A)<br\/> $\\Rightarrow CD \\perp AC \\Rightarrow \\widehat{ACD} = 90^{o} $ <br\/> X\u00e9t $\\triangle ABC $ v\u00e0 $\\triangle CDA $ c\u00f3: <br\/> + $AB = DC $ (ch\u1ee9ng minh tr\u00ean) <br\/> + $\\widehat{BAC} = \\widehat{DCA} (= 90^{o}) $ <br\/> + $AC$ c\u1ea1nh chung <br\/> Do \u0111\u00f3 $\\triangle ABC = \\triangle CDA (c.g.c) $ <br\/> $\\Rightarrow BC = DA $ <br\/> Do \u0111\u00f3: $AM = \\dfrac{1}{2} BC $ (v\u00ec $MA = \\dfrac{DA}{2} $)<br\/><span class='basic_pink'>\u0110\u00e1p \u00e1n \u0111\u00fang l\u00e0 D. <\/span>","column":2}]}],"id_ques":1786},{"time":24,"part":[{"time":3,"title":"N\u1ed1i t\u1eeb ho\u1eb7c c\u1ee5m t\u1eeb \u1edf c\u1ed9t b\u00ean tr\u00e1i v\u1edbi c\u1ed9t b\u00ean ph\u1ea3i \u0111\u1ec3 \u0111\u01b0\u1ee3c c\u00e2u ho\u00e0n ch\u1ec9nh","title_trans":"Bi\u1ebft r\u1eb1ng C l\u00e0 trung \u0111i\u1ec3m c\u1ee7a AD v\u00e0 $\\widehat{A} = \\widehat{D}$ <br\/> H\u00e3y n\u1ed1i c\u00e1c \u00fd v\u1edbi nhau \u0111\u1ec3 \u0111\u01b0\u1ee3c m\u1ed9t b\u00e0i ch\u1ee9ng minh $BC = EC$ ","audio":"","temp":"matching","correct":[["2","4","1","3"]],"list":[{"point":10,"ques":"","image":"https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai16/lv3/img\/H7B16-31.png","left":[" Theo b\u00e0i, C l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AD$ ","X\u00e9t hai tam gi\u00e1c $ABC$ v\u00e0 $DEC$ c\u00f3:","$\\Rightarrow \\triangle ABC = \\triangle DEC $","$\\Rightarrow BC = EC$"],"right":["theo tr\u01b0\u1eddng h\u1ee3p g\u00f3c-c\u1ea1nh-g\u00f3c","$\\Rightarrow AC=CD$","\u0111i\u1ec1u ph\u1ea3i ch\u1ee9ng minh","+ $\\widehat{A} = \\widehat{D}$; $AC=CD$ <br\/> $\\widehat{ACB} = \\widehat{DCE} $ (\u0111\u1ed1i \u0111\u1ec9nh)"],"top":100,"explain":"<span class='basic_left'> <b>B\u00e0i ch\u1ee9ng minh ho\u00e0n ch\u1ec9nh nh\u01b0 sau: <\/b> <br\/> Theo b\u00e0i: C l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AD$ <br\/> $\\Rightarrow AC=CD$ <br\/> X\u00e9t hai tam gi\u00e1c $ABC$ v\u00e0 $DEC$ c\u00f3: <br\/> + $\\widehat{A} = \\widehat{D}$ <br\/> + $AC=CD$ <br\/> + $\\widehat{ACB} = \\widehat{DCE} $ (\u0111\u1ed1i \u0111\u1ec9nh) <br\/> $\\Rightarrow \\triangle ABC = \\triangle DEC $ ( g- c - g) <br\/> $\\Rightarrow BC = EC$ (\u0111pcm) "}]}],"id_ques":1787},{"time":24,"part":[{"title":"\u0110i\u1ec1n t\u1eeb th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["120"]]],"list":[{"point":10,"width":20,"content":"","type_input":"","ques":"<span class='basic_left'> Cho tam gi\u00e1c $ABC$.<br\/> \u1ede ph\u00eda ngo\u00e0i tam gi\u00e1c $ABC$ v\u1ebd c\u00e1c tam gi\u00e1c \u0111\u1ec1u $ABD$ v\u00e0 $ACE$. <br\/>G\u1ecdi I l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $DC$ v\u00e0 $BE$. <br\/> S\u1ed1 \u0111o g\u00f3c $\\widehat{BIC} $ l\u00e0: $\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}^{o}$ ","explain":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai16/lv3/img\/H7B16-32.png' \/><\/center><span class='basic_left'>Ta c\u00f3 $\\triangle ADB $ \u0111\u1ec1u n\u00ean $\\widehat{DAB} = 60^{o} $, $\\triangle ACE $ \u0111\u1ec1u n\u00ean $\\widehat{CAE} = 60^{o}$ <br\/> $\\Rightarrow \\widehat{DAC} = \\widehat{BAE} (= 60^{o} + \\widehat{BAC}) $ <br\/> X\u00e9t $\\triangle ADC $ v\u00e0 $\\triangle ABE $ c\u00f3: <br\/> + $ AD = AB$ (gi\u1ea3 thi\u1ebft)<br\/> + $ \\widehat{DAC} = \\widehat{BAE}$<br\/> + $ AC = AE$ <br\/> Do \u0111\u00f3 $\\triangle ADC = \\triangle ABE (c.g.c) $ <br\/> $\\Rightarrow \\widehat{D_{1}} = \\widehat{B_{1}} $ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) <br\/> G\u1ecdi K l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $AB$ v\u00e0 $CD$. <br\/> Tam gi\u00e1c $KAD$ v\u00e0 $KIB$ c\u00f3: <br\/> + $\\widehat{D_{1}} = \\widehat{B_{1}} $ (ch\u1ee9ng minh tr\u00ean)<br\/> + $\\widehat{K_{1}} = \\widehat{K_{2}} $ (\u0111\u1ed1i \u0111\u1ec9nh) <br\/> $\\Rightarrow \\widehat{KAD} = \\widehat{KIB} $ <br\/> M\u00e0 $\\widehat{KAD} = 60^{o} \\Rightarrow \\widehat{KIB} = 60^{o} $ <br\/> Ta c\u00f3 $\\widehat{KIB} + \\widehat{BIC} = 180^{o} $ (k\u1ec1 b\u00f9) <br\/> $\\Rightarrow \\widehat{BIC} = 180^{o} - 60^{o} = 120^{o} $<br\/> <span class='basic_pink'> \u0110\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0: 120<\/span>"}]}],"id_ques":1788},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":" <b>T\u00ecm c\u00e2u tr\u1ea3 l\u1eddi sai:<\/b> <br\/> Cho h\u00ecnh v\u1ebd, bi\u1ebft: <br\/> $ AB = 1cm, BC = \\sqrt{2} cm , \\widehat{ADC} = 45^{o} $. Khi \u0111\u00f3: <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai16/lv3/img\/H7B16-33.png' \/><\/center> ","select":["A. $\\widehat{ABC} = \\widehat{ACB} = \\widehat{CAD} $","B. $\\widehat{BAC} = 90^{o}, \\widehat{BAD} = 135^{o} $","C. $AB \/\/ CD, AD \/\/ BC $ ","D. $AD = CD = 1cm $ "],"explain":" <span class='basic_left'> Ta c\u00f3 $\\triangle ABC $ vu\u00f4ng t\u1ea1i A n\u00ean theo \u0111\u1ecbnh l\u00ed Py-ta-go c\u00f3: <br\/> $BC^{2} = AB^{2} + AC^{2} \\\\ \\Rightarrow AC^{2} = BC^{2} - AB^{2} \\\\ = (\\sqrt{2})^{2} - (1)^{2} \\\\ = 1^{2} \\\\ \\Rightarrow AC = 1 cm $ <br\/> C\u00f3 $AB = AC = 1cm$ n\u00ean $\\triangle ABC $ vu\u00f4ng c\u00e2n t\u1ea1i A <br\/> Suy ra $\\widehat{ABC} = \\widehat{ACB} = \\widehat{CAD} (= 45^{o}) \\Rightarrow $ <b>A \u0111\u00fang <\/b> <br\/> Tam gi\u00e1c $ACD$ vu\u00f4ng t\u1ea1i C, c\u00f3 $\\widehat{CAD} = 45^{o} $ <br\/> $\\Rightarrow \\widehat{CAD} = 45^{o} $ <br\/> $\\Rightarrow \\widehat{BAD} = 90^{o} + 45^{o} = 135^{o} \\Rightarrow $ <b>B \u0111\u00fang<\/b> <br\/> C\u00f3 $\\widehat{BAC}= \\widehat{ACD} (= 90^{o}) \\Rightarrow AB \/\/ CD$ (so le trong) <br\/> C\u00f3 $\\widehat{CAD}= \\widehat{ACB} (= 45^{o}) \\Rightarrow AD \/\/ BC$ (so le trong) <br\/> $ \\Rightarrow $ <b>C \u0111\u00fang<\/b> <br\/> Tam gi\u00e1c $ACD$ vu\u00f4ng c\u00e2n t\u1ea1i C n\u00ean: <br\/> $CA = CD = 1cm $ (v\u00ec $AC = AB = 1cm$) <br\/> $AD^{2} = AC^{2} + CD^{2} = 1^{2} + 1^{1} = 2 \\\\ \\Rightarrow AD = \\sqrt{2} $ <br\/> $\\Rightarrow $ <b>D sai<\/b> <br\/> <br\/> <span class='basic_pink'> C\u00e2u tr\u1ea3 l\u1eddi sai l\u00e0 D. <\/span> ","column":2}]}],"id_ques":1789},{"time":24,"part":[{"time":3,"title":"S\u1eafp x\u1ebfp c\u00e1c c\u00e2u","title_trans":"<span class='basic_left'> Cho $\\triangle ABC $ c\u00e2n t\u1ea1i C. C\u00e1c \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $CA$ v\u00e0 $CB$ c\u1eaft nhau t\u1ea1i I. <br\/> H\u00e3y s\u1eafp x\u1ebfp c\u00e1c \u00fd ch\u1ee9ng minh $CI$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c C. ","temp":"sequence","correct":[[[4],[3],[5],[1],[2],[6]]],"list":[{"point":10,"image":"https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai16/lv3/img\/H7B16-34.png","left":["Do \u0111\u00f3 $\\triangle CEI = \\triangle CFI $ (c\u1ea1nh huy\u1ec1n - c\u1ea1nh g\u00f3c vu\u00f4ng)","X\u00e9t $\\triangle CEI $ v\u00e0 $\\triangle CFI $ c\u00f3: $CE = CF$ (ch\u1ee9ng minh tr\u00ean); c\u1ea1nh $CI$ chung","Suy ra $\\widehat{C_{1}} = \\widehat{C_{2}} $ ","Ta c\u00f3: $CA = CB $ ($\\triangle ABC $ c\u00e2n t\u1ea1i C), <br\/> $CE = \\dfrac{1}{2}CA, CF = \\dfrac{1}{2}CB $ (gi\u1ea3 thi\u1ebft)","Suy ra $CE = CF$","V\u1eady $CI$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c C. "],"top":80,"explain":"<span class='basic_left'> <b> B\u00e0i ch\u1ee9ng minh ho\u00e0n ch\u1ec9nh nh\u01b0 sau: <\/b> <br\/> Ta c\u00f3: $CA = CB $ ($\\triangle ABC $ c\u00e2n t\u1ea1i C), <br\/> $CE = \\dfrac{1}{2}CA, CF = \\dfrac{1}{2}CB $ (gi\u1ea3 thi\u1ebft) <br\/> Suy ra CE = CF <br\/> X\u00e9t hai tam gi\u00e1c vu\u00f4ng $\\triangle CEI $ v\u00e0 $\\triangle CFI $ c\u00f3: <br\/> + $CE = CF$ (ch\u1ee9ng minh tr\u00ean)<br\/> + c\u1ea1nh CI chung <br\/> Do \u0111\u00f3 $\\triangle CEI = \\triangle CFI $ (c\u1ea1nh huy\u1ec1n - c\u1ea1nh g\u00f3c vu\u00f4ng) <br\/>Suy ra $\\widehat{C_{1}} = \\widehat{C_{2}} $ <br\/> V\u1eady CI l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c C. "}]}],"id_ques":1790}],"lesson":{"save":0,"level":3}}