{"segment":[{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"\u0110\u1ec1 chung cho ba c\u00e2u","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":" <span class='basic_left'> Cho t\u1ee9 gi\u00e1c $ABCD$. G\u1ecdi $E, F, G, H$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AB, AC, DC, DB$. <br\/> <b> C\u00e2u 1:<\/b> T\u1ee9 gi\u00e1c $EFGH$ l\u00e0 h\u00ecnh g\u00ec? <\/span>","select":[" A. H\u00ecnh b\u00ecnh h\u00e0nh"," B. H\u00ecnh ch\u1eef nh\u1eadt","C. H\u00ecnh vu\u00f4ng","D. H\u00ecnh thoi"],"explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai8/lv3/img\/H818_K1_1.jpg' \/><\/center> X\u00e9t $\\Delta ABC$ c\u00f3: $AE=BE; AF=CF$ <br\/> $\\Rightarrow EF$ l\u00e0 \u0111\u01b0\u1eddng trung b\u00ecnh c\u1ee7a $\\Delta ABC$. <br\/> $\\Rightarrow EF \/\/BC$ v\u00e0 $EF= \\dfrac{1}{2}BC $. <br\/> X\u00e9t $\\Delta BCD$ c\u00f3: $DH=BH; DG=CG$ (gi\u1ea3 thi\u1ebft)<br\/>$\\Rightarrow GH$ l\u00e0 \u0111\u01b0\u1eddng trung b\u00ecnh c\u1ee7a tam gi\u00e1c $BCD$ <br\/> $\\Rightarrow GH \/\/BC$ v\u00e0 $GH= \\dfrac{1}{2}BC $ <br\/> Do \u0111\u00f3: $EF\/\/GH$ v\u00e0 $EF=GH$ <br\/> $\\Rightarrow EFGH$ l\u00e0 h\u00ecnh b\u00ecnh h\u00e0nh (d\u1ea5u hi\u1ec7u nh\u1eadn bi\u1ebft). <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A. <\/span><\/span> ","column":2}]}],"id_ques":1511},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"\u0110\u1ec1 chung cho ba c\u00e2u","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":" <span class='basic_left'> Cho t\u1ee9 gi\u00e1c $ABCD$. G\u1ecdi $E, F, G, H$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AB, AC, DC, DB$. <br\/> <b> C\u00e2u 2:<\/b> N\u1ebfu $AD\\bot BC$ th\u00ec $EFGH$ l\u00e0 h\u00ecnh g\u00ec? <\/span>","select":[" A. H\u00ecnh b\u00ecnh h\u00e0nh"," B. H\u00ecnh ch\u1eef nh\u1eadt","C. H\u00ecnh vu\u00f4ng","D. H\u00ecnh thoi"],"explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai8/lv3/img\/H818_K1_1a.png' \/><\/center> X\u00e9t $\\Delta ABD$ c\u00f3: $ AE=EB; BH=DH$ (gi\u1ea3 thi\u1ebft) <br\/> $\\Rightarrow EH$ l\u00e0 \u0111\u01b0\u1eddng trung b\u00ecnh c\u1ee7a tam gi\u00e1c $ABD$ <br\/> $\\Rightarrow EH \/\/AD$ <br\/> L\u1ea1i c\u00f3: $EF\/\/BC$ (ch\u1ee9ng minh c\u00e2u 1) <br\/> M\u00e0 $AD\\bot BC$ n\u00ean $EH \\bot EF$ t\u1ea1i $E$. <br\/>Theo c\u00e2u 1, ta \u0111\u00e3 ch\u1ee9ng minh \u0111\u01b0\u1ee3c $EFGH$ l\u00e0 h\u00ecnh b\u00ecnh h\u00e0nh. <br\/> Do \u0111\u00f3 $EFGH$ l\u00e0 h\u00ecnh ch\u1eef nh\u1eadt (d\u1ea5u hi\u1ec7u nh\u1eadn bi\u1ebft). <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B. <\/span><\/span> ","column":2}]}],"id_ques":1512},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"\u0110\u1ec1 chung cho ba c\u00e2u","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":" <span class='basic_left'> Cho t\u1ee9 gi\u00e1c $ABCD$. G\u1ecdi $E, F, G, H$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AB, AC, DC, DB$. <br\/> <b> C\u00e2u 3:<\/b> \u0110\u1ec3 $EFGH$ l\u00e0 h\u00ecnh vu\u00f4ng th\u00ec t\u1ee9 gi\u00e1c $ABCD$ c\u1ea7n th\u00eam \u0111i\u1ec1u ki\u1ec7n l\u00e0: <\/span>","select":[" A. $AD\\bot BC; AD=BC$"," B. $AD\\bot BC$","C. $AD=BC$"],"explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai8/lv3/img\/H818_K1_1.jpg' \/><\/center> Theo c\u00e2u 1, ta \u0111\u00e3 ch\u1ee9ng minh \u0111\u01b0\u1ee3c $EFGH$ l\u00e0 h\u00ecnh b\u00ecnh h\u00e0nh <br\/> \u0110\u1ec3 $EFGH$ l\u00e0 h\u00ecnh vu\u00f4ng th\u00ec c\u1ea7n $EEGH$ l\u00e0 h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 $EH=EF$ <br\/> M\u00e0 $EH=\\dfrac{1}{2}AD;EF= \\dfrac{1}{2} BC$ (theo c\u00e2u 2) <br\/> $\\Rightarrow AD=BC$ (1) <br\/> Theo c\u00e2u 2: $EFGH$ l\u00e0 h\u00ecnh ch\u1eef nh\u1eadt th\u00ec $ABCD$ c\u1ea7n th\u00eam \u0111i\u1ec1u ki\u1ec7n $AD\\bot BC$ (2) <br\/> T\u1eeb (1) v\u00e0 (2) suy ra t\u1ee9 gi\u00e1c $EFGH$ l\u00e0 h\u00ecnh vu\u00f4ng th\u00ec t\u1ee9 gi\u00e1c $ABCD$ ph\u1ea3i th\u1ecfa m\u00e3n: $AD\\bot BC$ v\u00e0 $AD=BC$. <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A. <\/span><\/span> ","column":3}]}],"id_ques":1513},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":" <span class='basic_left'> Cho h\u00ecnh vu\u00f4ng $ABCD$. \u0110i\u1ec3m $E$ thu\u1ed9c mi\u1ec1n trong h\u00ecnh vu\u00f4ng sao cho $\\widehat{EAB}=\\widehat{EBA}=15^o$. D\u1ef1ng tam gi\u00e1c \u0111\u1ec1u $FEB$ sao cho $F$ v\u00e0 $C$ \u1edf c\u00f9ng ph\u00eda v\u1edbi $EB$. <br\/> <b> C\u00e2u 1:<\/b> Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t \u0111i\u1ec1n v\u00e0o ch\u1ed7 tr\u1ed1ng sau: <br\/> Ta ch\u1ee9ng minh \u0111\u01b0\u1ee3c $\\Delta ABE = $ ___________ <\/span>","select":[" A. $\\Delta FCB$"," B. $\\Delta CBF$","C. $\\Delta BEC$"],"explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai8/lv3/img\/H818_K1_2.jpg' \/><\/center> Ta c\u00f3 $\\widehat{ABC}=90^o$ <br\/> Suy ra: $\\widehat{ABE}+\\widehat{EBF}+\\widehat{FBC}=90^o$ <br\/> $\\Rightarrow 15^o+60^o+\\widehat{FBC}=90^o$ <br\/> $\\Rightarrow \\widehat{FBC}=15^o$ <br\/> X\u00e9t $\\Delta ABE$ v\u00e0 $\\Delta CBF$ c\u00f3: <br\/> + $AB=BC$ (gi\u1ea3 thi\u1ebft) <br\/> + $\\widehat{EBA}=\\widehat{FBC}=15^o$ (ch\u1ee9ng minh tr\u00ean) <br\/> + $BE=BF$ (gi\u1ea3 thi\u1ebft) <br\/> $\\Rightarrow \\Delta ABE=\\Delta CBF$ (c - g - c). <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B. <\/span><\/span> ","column":3}]}],"id_ques":1514},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":" <span class='basic_left'> Cho h\u00ecnh vu\u00f4ng $ABCD$. \u0110i\u1ec3m $E$ thu\u1ed9c mi\u1ec1n trong h\u00ecnh vu\u00f4ng sao cho $\\widehat{EAB}=\\widehat{EBA}=15^o$. D\u1ef1ng tam gi\u00e1c \u0111\u1ec1u $FEB$ sao cho $F$ v\u00e0 $C$ \u1edf c\u00f9ng ph\u00eda v\u1edbi $EB$. <br\/> <b> C\u00e2u 2:<\/b> $\\Delta CBF$ l\u00e0 tam gi\u00e1c g\u00ec? <\/span>","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/lop8/toan/hinhhoc/bai8/lv3/img\/H818_K1_2.jpg' \/><\/center> Theo c\u00e2u 1, ta ch\u1ee9ng minh \u0111\u01b0\u1ee3c $ \\Delta ABE=\\Delta CBF$ (c - g - c) <br\/> M\u00e0 $\\Delta ABE$ c\u00e2n t\u1ea1i $E$ (gi\u1ea3 thi\u1ebft) <br\/> $\\Rightarrow \\Delta CBF$ c\u00e2n t\u1ea1i $F$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A. <\/span><\/span> ","column":2}]}],"id_ques":1515},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"fill_the_blank","correct":[[["15"]]],"list":[{"point":10,"width":40,"type_input":"","ques":"<span class='basic_left'> Cho h\u00ecnh vu\u00f4ng $ABCD$. \u0110i\u1ec3m E thu\u1ed9c mi\u1ec1n trong h\u00ecnh vu\u00f4ng sao cho $\\widehat{EAB}=\\widehat{EBA}=15^o$. D\u1ef1ng tam gi\u00e1c \u0111\u1ec1u $FEB$ sao cho $F$ v\u00e0 $C$ \u1edf c\u00f9ng ph\u00eda v\u1edbi $EB$. <br\/> <b> C\u00e2u 3:<\/b> T\u00ednh s\u1ed1 \u0111o $\\widehat {CEF}$. <br\/><br\/> <b> \u0110\u00e1p \u00e1n: <\/b> $\\widehat{CEF}=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}^o$ <\/span> ","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai8/lv3/img\/H818_K1_2.jpg' \/><\/center> Theo gi\u1ea3 thi\u1ebft $EF=BF$ <br\/> Theo c\u00e2u 2, ta ch\u1ee9ng minh \u0111\u01b0\u1ee3c: $\\Delta CBF$ c\u00e2n t\u1ea1i $F$ n\u00ean $BF=CF$ <br\/> Do \u0111\u00f3 $EF=CF$ $\\Rightarrow \\Delta EFC$ c\u00e2n t\u1ea1i $F$ <br\/> M\u00e0 $\\widehat{CFE}=360^o-\\widehat{BFE}-\\widehat{BFC}$ <br\/> $=360^o-60^o-(180^o-2.15^o)$ <br\/> $=150^o$ <br\/> $\\Rightarrow \\widehat{CEF}=\\dfrac{180^o-150^o}{2}=15^o$ <br\/> <span class='basic_pink'> V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $15$. <\/span> <\/span> "}]}],"id_ques":1516},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":" <span class='basic_left'> Cho h\u00ecnh vu\u00f4ng $ABCD$. \u0110i\u1ec3m $E$ thu\u1ed9c mi\u1ec1n trong h\u00ecnh vu\u00f4ng sao cho $\\widehat{EAB}=\\widehat{EBA}=15^o$. D\u1ef1ng tam gi\u00e1c \u0111\u1ec1u $FEB$ sao cho $F$ v\u00e0 $C$ \u1edf c\u00f9ng ph\u00eda v\u1edbi $EB$. <br\/> <b> C\u00e2u 4:<\/b> $\\Delta CDE$ l\u00e0 tam gi\u00e1c g\u00ec? <\/span>","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/lop8/toan/hinhhoc/bai8/lv3/img\/H818_K1_2.jpg' \/><\/center> Ta d\u1ec5 d\u00e0ng ch\u1ee9ng minh \u0111\u01b0\u1ee3c: $\\Delta ABE=\\Delta ECF$ (c - g - c) <br\/> $\\Rightarrow AB=EC$ <br\/> M\u00e0 $AB=DC\\Rightarrow DC=EC$ (1) <br\/> M\u1eb7t kh\u00e1c: $\\widehat{ECD}=90^o-\\widehat{FCB}-\\widehat{FCE}$ <br\/> $=90^o-15^o-15^o$<br\/> $=60^o$ (2) <br\/> T\u1eeb (1) v\u00e0 (2) suy ra $\\Delta CED$ \u0111\u1ec1u. <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B. <\/span><\/span> ","column":2}]}],"id_ques":1517},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":" <span class='basic_left'> Cho tam gi\u00e1c $ABC$ vu\u00f4ng c\u00e2n t\u1ea1i $A$ c\u00f3 $BC=2a$ ($a > 0$) <br\/> <b> C\u00e2u 1:<\/b> \u0110\u1ed9 d\u00e0i \u0111\u01b0\u1eddng cao $AD$ c\u1ee7a tam gi\u00e1c $ABC$ t\u00ednh theo $a$ l\u00e0: <\/span>","select":[" A. $\\dfrac{a}{2}$ (\u0111\u01a1n v\u1ecb \u0111\u1ed9 d\u00e0i)"," B. $a$ (\u0111\u01a1n v\u1ecb \u0111\u1ed9 d\u00e0i)","C. $\\dfrac{3a}{2}$ (\u0111\u01a1n v\u1ecb \u0111\u1ed9 d\u00e0i)"],"hint":"T\u00ednh $BC$ r\u1ed3i x\u00e9t trong tam gi\u00e1c $ADB$ \u0111\u1ec3 t\u00ednh $AD$.","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai8/lv3/img\/H818_K1_3.jpg' \/><\/center> \u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pi- ta - go v\u00e0o tam gi\u00e1c vu\u00f4ng c\u00e2n $ABC$ ta c\u00f3: <br\/> $AB^2+AC^2=BC^2$ <br\/> $\\Rightarrow 2AB^2=4a^2$ <br\/> $\\Rightarrow AB^2=2a^2$ <br\/> $\\Rightarrow AB=a\\sqrt{2}$ <br\/> Trong tam gi\u00e1c vu\u00f4ng c\u00e2n $ABC$, c\u00f3 $AD$ l\u00e0 \u0111\u01b0\u1eddng cao \u0111\u1ed3ng th\u1eddi c\u0169ng l\u00e0 \u0111\u01b0\u1eddng trung tuy\u1ebfn. <br\/> $\\Rightarrow BD=\\dfrac{1}{2}BC=\\dfrac{2a}{2}=a$ <br\/> X\u00e9t $\\Delta ADB$ c\u00f3: <br\/> $AB^2=AD^2+BD^2$ (\u0111\u1ecbnh l\u00ed Py - ta - go) <br\/> $\\Rightarrow 2a^2=AD^2+a^2$ <br\/> $\\Rightarrow AD^2=a^2$ <br\/> $\\Rightarrow AD=a$ (\u0111\u01a1n v\u1ecb \u0111\u1ed9 d\u00e0i). <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B. <\/span><\/span> ","column":3}]}],"id_ques":1518},{"time":24,"part":[{"title":"L\u1ef1a 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 tam gi\u00e1c $ABC$ vu\u00f4ng c\u00e2n t\u1ea1i $A$ c\u00f3 $BC=2a$ ($a > 0$). <br\/> <b> C\u00e2u 2:<\/b> G\u1ecdi $M$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AC$. Tr\u00ean tia $DM$ l\u1ea5y \u0111i\u1ec3m $E$ sao cho $DM=ME$. Di\u1ec7n t\u00edch c\u1ee7a $ADCE$ l\u00e0: <\/span>","select":[" A. $a^2$ (\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch)"," B. $a$ (\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch)","C. $2a$ (\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch)"],"hint":"X\u00e9t xem $ADCE$ l\u00e0 h\u00ecnh g\u00ec r\u1ed3i t\u00ednh di\u1ec7n t\u00edch c\u1ee7a n\u00f3.","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai8/lv3/img\/H818_K1_3.jpg' \/><\/center> Do $ DM=EM; AM=CM$ n\u00ean $ADCE$ l\u00e0 h\u00ecnh b\u00ecnh h\u00e0nh (hai \u0111\u01b0\u1eddng ch\u00e9o c\u1eaft nhau t\u1ea1i trung \u0111i\u1ec3m c\u1ee7a m\u1ed7i \u0111\u01b0\u1eddng) <br\/> M\u00e0 $\\widehat{ADC}=90^o$ n\u00ean $ADCE$ l\u00e0 h\u00ecnh ch\u1eef nh\u1eadt (d\u1ea5u hi\u1ec7u nh\u1eadn bi\u1ebft) <br\/> L\u1ea1i c\u00f3: $AD=DC=a$ <br\/> $\\Rightarrow ADCE$ l\u00e0 h\u00ecnh vu\u00f4ng (d\u1ea5u hi\u1ec7u nh\u1eadn bi\u1ebft) <br\/> Di\u1ec7n t\u00edch $ADCE$ l\u00e0: <br\/> $AD^2=a^2$ (\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch) <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A. <\/span><\/span> ","column":3}]}],"id_ques":1519},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":" <span class='basic_left'> Cho $\\Delta$ $ABC$ c\u00e2n t\u1ea1i $A$, c\u00e1c \u0111\u01b0\u1eddng trung tuy\u1ebfn $BD$ v\u00e0 $CE$ c\u1eaft nhau \u1edf $G$. G\u1ecdi $H, K$ theo th\u1ee9 t\u1ef1 l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $GB, GC$. <br\/> Tam gi\u00e1c c\u00e2n $ABC$ c\u00f3 th\u00eam \u0111i\u1ec1u ki\u1ec7n g\u00ec \u0111\u1ec3 $DEHK$ l\u00e0 h\u00ecnh vu\u00f4ng? <\/span>","select":["A. $\\widehat{A}=90^o$","B. $CE \\bot BD$","C. $\\widehat{B}=60^o$"],"hint":"X\u00e1c \u0111\u1ecbnh $DEHK$ \u0111\u00e3 cho l\u00e0 h\u00ecnh g\u00ec? <br\/> T\u00ecm \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 $DEHK$ l\u00e0 h\u00ecnh vu\u00f4ng.","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai8/lv3/img\/H818_K1_4.jpg' \/><\/center>Trong $\\Delta$ $ABC$ c\u00f3: $AE = EB; AD = DC$ (gi\u1ea3 thi\u1ebft) n\u00ean $ED\/\/BC;\\,\\,ED=\\dfrac{1}{2}BC$ (1)<br\/> Trong $\\Delta BGC$ c\u00f3: $HG = HB; KG = KC$ (gi\u1ea3 thi\u1ebft) n\u00ean $HK\/\/BC;HK=\\dfrac{1}{2}BC$ (2) <br\/> T\u1eeb (1) v\u00e0 (2) suy ra $ED \/\/ HK$ v\u00e0 $ED = HK$ <br\/> Do \u0111\u00f3 $DEHK$ l\u00e0 h\u00ecnh b\u00ecnh h\u00e0nh. (d\u1ea5u hi\u1ec7u nh\u1eadn bi\u1ebft) (3) <br\/> M\u00e0 $EK=EG+GK=\\dfrac{1}{3}CE+\\dfrac{1}{3}CE=\\dfrac{2}{3}CE$ ( t\u00ednh ch\u1ea5t ba \u0111\u01b0\u1eddng trung tuy\u1ebfn trong tam gi\u00e1c) <br\/> $HD=GD+HG=\\dfrac{1}{3}BD+\\dfrac{1}{3}BD=\\dfrac{2}{3}BD$ ( t\u00ednh ch\u1ea5t ba \u0111\u01b0\u1eddng trung tuy\u1ebfn trong tam gi\u00e1c) <br\/> Do $\\Delta$ $ABC$ c\u00e2n t\u1ea1i $A$ n\u00ean $CE = BD$, suy ra $EK = DH$ (4) <br\/> T\u1eeb (3) v\u00e0 (4) suy ra $DEHK$ l\u00e0 h\u00ecnh ch\u1eef nh\u1eadt (d\u1ea5u hi\u1ec7u nh\u1eadn bi\u1ebft). <br\/> Mu\u1ed1n $DEHK$ l\u00e0 h\u00ecnh vu\u00f4ng th\u00ec $EK\\bot DH\\Rightarrow CE\\bot BD$ <br\/> V\u1eady $\\Delta$ $ABC$ c\u00f3 $CE\\bot BD$ th\u00ec $DEHK$ l\u00e0 h\u00ecnh vu\u00f4ng.<br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0: B <\/span><\/span> ","column":3}]}],"id_ques":1520}],"lesson":{"save":0,"level":3}}