Tính chất đường phân giác của tam giác

Home Lớp 8 Toán lớp 8 - Sách Kết nối tri thức Bài 17 - Tính chất đường phân giác của tam giácBài tập nâng cao

{"common":{"save":0,"post_id":"6585","level":3,"total":10,"point":10,"point_extra":0},"segment":[{"id":"6243","post_id":"6585","mon_id":"1158921","chapter_id":"1158932","question":"Cho tam giác ABC cân \u1edf A có AC = 7,5 cm, BC = 5 cm. \u0110\u01b0\u1eddng phân giác c\u1ee7a góc C c\u1eaft AB t\u1ea1i D. Khi \u0111ó \u0111\u1ed9 dài \u0111o\u1ea1n th\u1eb3ng AD là:<\/p><svg height=\"260\" width=\"270\"> <g><title><\/title><rect fill=\"#fff\" height=\"262\" id=\"canvas_background\" width=\"272\" x=\"-1\" y=\"-1\"><\/rect> <g display=\"none\" id=\"canvasGrid\"> <rect fill=\"url(#gridpattern)\" height=\"100%\" id=\"svg_7\" stroke-width=\"0\" width=\"100%\" x=\"0\" y=\"0\"><\/rect> <\/g> <\/g> <g><title><\/title><path d=\"m28.66975,234.86506l102.5,-211.00001l102.5,211.00001l-204.99999,0l-0.00001,0z\" fill=\"#d4aaff\" id=\"svg_1\" stroke=\"#000\" stroke-width=\"1.5\"><\/path> <text fill=\"#000000\" fill-opacity=\"null\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"24\" id=\"svg_2\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"0\" text-anchor=\"start\" x=\"122.95453\" xml:space=\"preserve\" y=\"20.88636\">A<\/text> <text fill=\"#000000\" fill-opacity=\"null\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"24\" id=\"svg_3\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"0\" text-anchor=\"start\" x=\"6.95453\" xml:space=\"preserve\" y=\"240.88635\">B<\/text> <text fill=\"#000000\" fill-opacity=\"null\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"24\" id=\"svg_4\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"0\" text-anchor=\"start\" x=\"239.95453\" xml:space=\"preserve\" y=\"237.88635\">C<\/text> <text fill=\"#000000\" fill-opacity=\"null\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"24\" id=\"svg_5\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"0\" text-anchor=\"start\" x=\"40.95453\" xml:space=\"preserve\" y=\"152.88635\">D<\/text> <line fill=\"none\" fill-opacity=\"null\" id=\"svg_6\" stroke=\"#000\" stroke-linecap=\"null\" stroke-linejoin=\"null\" stroke-opacity=\"null\" stroke-width=\"1.5\" x1=\"231.95453\" x2=\"70.95453\" y1=\"233.88635\" y2=\"144.88635\"><\/line> <text fill=\"none\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"17\" font-style=\"normal\" font-weight=\"normal\" id=\"svg_8\" stroke=\"#000\" stroke-width=\"1.5\" text-anchor=\"start\" x=\"193.95453\" xml:space=\"preserve\" y=\"138.88635\">7,5 cm<\/text> <text fill=\"none\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"17\" id=\"svg_9\" stroke=\"#000\" stroke-width=\"2\" text-anchor=\"start\" transform=\"rotate(41.237606048583984 215.28964233398432,211.83807373046878) matrix(0.4709177017211914,0,0,1,138.59550414659316,0) \" x=\"160.02589\" xml:space=\"preserve\" y=\"217.88635\">(<\/text> <text fill=\"none\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"17\" id=\"svg_11\" stroke=\"#000\" stroke-width=\"2\" text-anchor=\"start\" transform=\"rotate(18.318695068359375 200.28964233398466,224.83807373046872) matrix(0.4709177017211914,0,0,1,138.59550414659316,0) \" x=\"128.17319\" xml:space=\"preserve\" y=\"230.88635\">(<\/text> <text fill=\"#000000\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"17\" id=\"svg_10\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"0\" text-anchor=\"start\" x=\"107.95453\" xml:space=\"preserve\" y=\"249.88635\">5 cm<\/text> <\/g> <\/svg><\/span><\/p>","options":["A.<\/strong> AD = 3 cm","B.<\/strong> AD = 3,5 cm","C.<\/strong> AD = 4,5 cm","D.<\/strong> AD = 4,2 cm"],"correct":"3","level":"3","hint":"","answer":"Vì tam giác ABC cân \u1edf A nên AB<\/strong><\/span> = AC = 7,5 cm<\/strong><\/span>.<\/p>Vì CD là \u0111\u01b0\u1eddng phân giác c\u1ee7a góc ACB và D∈AB nên theo tính ch\u1ea5t \u0111\u01b0\u1eddng phân giác trong tam giác, ta có: <\/p>$\\dfrac{AD}{BD}=\\dfrac{AC}{BC}=\\dfrac{7,5}{5}=\\dfrac{3}{2}$<\/span> hay $\\dfrac{AD}{BD}=\\dfrac{3}{2}$<\/span><\/p>Do \u0111ó $\\dfrac{AD}{3}=\\dfrac{BD}{2}=\\dfrac{AD+BD}{3+2}=\\dfrac{AB}{5} =\\dfrac{7,5}{5}$<\/span> (tính ch\u1ea5t dãy t\u1ec9 s\u1ed1 b\u1eb1ng nhau)<\/p>Suy ra AD = $\\dfrac{3\\space . \\space 7,5}{5}$<\/span> = 4,5 (cm)<\/p>Ch\u1ecdn C.<\/strong> AD = 4,5 cm<\/span><\/p>","type":"choose","extra_type":"classic","time":"0","user_id":"131","test":"0","date":"2023-08-06 04:31:09","option_type":"txt","len":1},{"id":"6244","post_id":"6585","mon_id":"1158921","chapter_id":"1158932","question":"Cho tam giác MNP có ND là \u0111\u01b0\u1eddng phân giác c\u1ee7a góc MNP (D∈MP) và $\\dfrac{MD}{MP}=\\dfrac{2}{7}$<\/span>. Tính \u0111\u1ed9 dài \u0111o\u1ea1n th\u1eb3ng NP.<\/p><svg height=\"200\" width=\"300\"> <g><title><\/title><rect fill=\"#fff\" height=\"202\" id=\"canvas_background\" width=\"302\" x=\"-1\" y=\"-1\"><\/rect> <g display=\"none\" height=\"100%\" id=\"canvasGrid\" overflow=\"visible\" width=\"100%\" x=\"0\" y=\"0\"> <rect fill=\"url(#gridpattern)\" height=\"100%\" stroke-width=\"0\" width=\"100%\" x=\"0\" y=\"0\"><\/rect> <\/g> <\/g> <g><title><\/title><path d=\"m90.64975,25.71504l181.99998,146l-242.99998,1z\" fill=\"#ffff00\" id=\"svg_1\" stroke=\"#000\" stroke-width=\"1.5\"><\/path> <text fill=\"#000000\" fill-opacity=\"null\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"24\" id=\"svg_2\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"0\" text-anchor=\"start\" x=\"79.95453\" xml:space=\"preserve\" y=\"21.88636\">M<\/text> <text fill=\"#000000\" fill-opacity=\"null\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"24\" id=\"svg_3\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"0\" text-anchor=\"start\" x=\"10.95453\" xml:space=\"preserve\" y=\"181.88635\">P<\/text> <text fill=\"#000000\" fill-opacity=\"null\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"24\" id=\"svg_4\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"0\" text-anchor=\"start\" x=\"274.95452\" xml:space=\"preserve\" y=\"178.88635\">N<\/text> <text fill=\"#000000\" fill-opacity=\"null\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"24\" id=\"svg_5\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"0\" text-anchor=\"start\" x=\"33.95453\" xml:space=\"preserve\" y=\"97.88636\">D<\/text> <line fill=\"none\" fill-opacity=\"null\" id=\"svg_6\" stroke=\"#000\" stroke-linecap=\"null\" stroke-linejoin=\"null\" stroke-opacity=\"null\" stroke-width=\"1.5\" x1=\"270.95452\" x2=\"55.95453\" y1=\"170.88635\" y2=\"106.88636\"><\/line> <text fill=\"#000000\" fill-opacity=\"null\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"24\" id=\"svg_7\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"0\" text-anchor=\"start\" transform=\"rotate(37.052207946777344 222.95452880859384,142.5227355957031) \" x=\"218.95453\" xml:space=\"preserve\" y=\"150.88635\">(<\/text> <text fill=\"#000000\" fill-opacity=\"null\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"24\" id=\"svg_8\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"0\" text-anchor=\"start\" transform=\"rotate(8.661446571350098 199.95545959472653,159.53410339355477) matrix(1.2500978708267212,0,0,1,-49.25790830470942,0) \" x=\"195.35465\" xml:space=\"preserve\" y=\"167.88635\">(<\/text> <text fill=\"#000000\" fill-opacity=\"null\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"17\" id=\"svg_9\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"0\" text-anchor=\"start\" x=\"164.95453\" xml:space=\"preserve\" y=\"80.88635\">14 cm<\/text> <\/g> <\/svg><\/span><\/p>","options":["A.<\/strong> NP = 35 cm","B.<\/strong> NP = 32 cm","C.<\/strong> NP = 45 cm","D.<\/strong> NP = 30 cm"],"correct":"1","level":"3","hint":"","answer":"Theo tính ch\u1ea5t c\u1ee7a dãy t\u1ec9 s\u1ed1 b\u1eb1ng nhau ta có:<\/p>$\\dfrac{MD}{MP}=\\dfrac{2}{7}$<\/span><\/p>suy ra $\\dfrac{MD}{MP-MD}=\\dfrac{2}{7-2}$<\/span><\/p>suy ra $ \\dfrac{MD}{PD}=\\dfrac{2}{5}$<\/span><\/p>Vì ND là \u0111\u01b0\u1eddng phân giác c\u1ee7a góc MNP và D∈MP nên theo tính ch\u1ea5t \u0111\u01b0\u1eddng phân giác trong tam giác ta có: <\/p>$\\dfrac{MN}{PN}=\\dfrac{MD}{PD}$<\/span> <\/p>thay s\u1ed1 $\\dfrac{14}{PN} =\\dfrac{2}{5} $<\/span><\/p>suy ra $PN = \\dfrac{14\\space .\\space 5}{2}=35$<\/span> (cm)<\/p>Ch\u1ecdn A.<\/strong> NP = 35 cm<\/span><\/p>","type":"choose","extra_type":"classic","time":"0","user_id":"131","test":"0","date":"2023-08-06 04:39:48","option_type":"txt","len":1},{"id":"6246","post_id":"6585","mon_id":"1158921","chapter_id":"1158932","question":"Cho tam giác ABC cân \u1edf A có BD và CE l\u1ea7n l\u01b0\u1ee3t là các \u0111\u01b0\u1eddng phân giác c\u1ee7a các góc ABC và góc ACB. Bi\u1ebft $\\dfrac{AE}{BE}=\\dfrac{3}{2}$<\/span>, tính t\u1ec9 s\u1ed1 $\\dfrac{CD}{AD}$<\/span>.<\/p><svg height=\"250\" width=\"260\"> <g><title><\/title><rect fill=\"#fff\" height=\"252\" id=\"canvas_background\" width=\"262\" x=\"-1\" y=\"-1\"><\/rect> <g display=\"none\" height=\"100%\" id=\"canvasGrid\" overflow=\"visible\" width=\"100%\" x=\"0\" y=\"0\"> <rect fill=\"url(#gridpattern)\" height=\"100%\" stroke-width=\"0\" width=\"100%\" x=\"0\" y=\"0\"><\/rect> <\/g> <\/g> <g><title><\/title><g id=\"svg_13\"> <path d=\"m22.15477,226.79505l108,-199.99999l108,199.99999l-215.99999,0z\" fill=\"#56ffff\" id=\"svg_1\" stroke=\"#000\" stroke-width=\"1.5\"><\/path> <line fill=\"none\" fill-opacity=\"null\" id=\"svg_2\" stroke=\"#000\" stroke-linecap=\"null\" stroke-linejoin=\"null\" stroke-opacity=\"null\" stroke-width=\"1.5\" x1=\"23.95453\" x2=\"193.95453\" y1=\"225.88635\" y2=\"141.88635\"><\/line> <line fill=\"none\" fill-opacity=\"null\" id=\"svg_3\" stroke=\"#000\" stroke-linecap=\"null\" stroke-linejoin=\"null\" stroke-opacity=\"null\" stroke-width=\"1.5\" x1=\"238.95452\" x2=\"67.95453\" y1=\"226.88635\" y2=\"139.88635\"><\/line> <text fill=\"#000000\" fill-opacity=\"null\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"24\" id=\"svg_4\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"0\" text-anchor=\"start\" transform=\"rotate(-33.46696090698242 47.95452880859369,198.52273559570312) \" x=\"43.95453\" xml:space=\"preserve\" y=\"206.88635\">)<\/text> <text fill=\"#000000\" fill-opacity=\"null\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"24\" id=\"svg_5\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"0\" text-anchor=\"start\" transform=\"rotate(-16.55707359313965 70.95452880859374,212.52273559570307) \" x=\"66.95453\" xml:space=\"preserve\" y=\"220.88635\">)<\/text> <text fill=\"#000000\" fill-opacity=\"null\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"24\" id=\"svg_6\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"0\" text-anchor=\"start\" transform=\"rotate(20.8510799407959 193.95452880859392,213.52273559570315) \" x=\"189.95453\" xml:space=\"preserve\" y=\"221.88635\">(<\/text> <text fill=\"#000000\" fill-opacity=\"null\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"24\" id=\"svg_7\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"0\" text-anchor=\"start\" transform=\"rotate(40.39131546020508 210.95452880859375,196.52273559570312) \" x=\"206.95453\" xml:space=\"preserve\" y=\"204.88635\">(<\/text> <text fill=\"#000000\" fill-opacity=\"null\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"24\" id=\"svg_8\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"0\" text-anchor=\"start\" x=\"118.95453\" xml:space=\"preserve\" y=\"21.88636\">A<\/text> <text fill=\"#000000\" fill-opacity=\"null\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"24\" id=\"svg_9\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"0\" text-anchor=\"start\" x=\"0.95453\" xml:space=\"preserve\" y=\"234.88636\">B<\/text> <text fill=\"#000000\" fill-opacity=\"null\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"24\" id=\"svg_10\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"0\" text-anchor=\"start\" x=\"240.95452\" xml:space=\"preserve\" y=\"229.88635\">C<\/text> <text fill=\"#000000\" fill-opacity=\"null\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"24\" id=\"svg_11\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"0\" text-anchor=\"start\" x=\"47.95453\" xml:space=\"preserve\" y=\"138.88636\">E<\/text> <text fill=\"#000000\" fill-opacity=\"null\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"24\" id=\"svg_12\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"0\" text-anchor=\"start\" x=\"198.95453\" xml:space=\"preserve\" y=\"139.88635\">D<\/text> <\/g> <\/g> <\/svg><\/span><\/p>","options":["A.<\/strong> $\\dfrac{CD}{AD}=\\dfrac{3}{2}$<\/span>","B.<\/strong> $\\dfrac{CD}{AD}=\\dfrac{2}{3}$<\/span>","C.<\/strong> $\\dfrac{CD}{AD}=\\dfrac{2}{1}$<\/span>","D.<\/strong> $\\dfrac{CD}{AD}=\\dfrac{1}{3}$<\/span>"],"correct":"2","level":"3","hint":"","answer":"Vì tam giác ABC cân \u1edf A nên AB = AC (1).<\/p>L\u1ea1i vì BD là \u0111\u01b0\u1eddng phân giác c\u1ee7a góc ABC nên theo tính ch\u1ea5t \u0111\u01b0\u1eddng phân giác ta có: $\\dfrac{CD}{AD}=\\dfrac{CB}{AB}$<\/span> (2)<\/p>L\u1ea1i vì CE là \u0111\u01b0\u1eddng phân giác c\u1ee7a góc ACB nên theo tính ch\u1ea5t \u0111\u01b0\u1eddng phân giác ta có: $\\dfrac{BE}{AE}=\\dfrac{BC}{AC}$<\/span> (3)<\/p>T\u1eeb (1), (2) và (3) suy ra $\\dfrac{CD}{AD}=\\dfrac{CB}{AB}$<\/span> = $\\dfrac{BE}{AE}=\\dfrac{BC}{AC}$<\/span> hay $\\dfrac{CD}{AD}=\\dfrac{BE}{AE}=\\dfrac{2}{3}$<\/span>.<\/p>Ch\u1ecdn B.<\/strong> $\\dfrac{CD}{AD}=\\dfrac{2}{3}$<\/span><\/span><\/p>","type":"choose","extra_type":"classic","time":"0","user_id":"131","test":"0","date":"2023-08-06 04:51:08","option_type":"math","len":0}]}