{"id":3877,"date":"2024-03-03T12:33:00","date_gmt":"2024-03-03T16:33:00","guid":{"rendered":"http:\/\/mathfun4kids.com\/mlog\/?p=3877"},"modified":"2024-10-25T12:32:26","modified_gmt":"2024-10-25T16:32:26","slug":"geometry-challenge-15","status":"publish","type":"post","link":"https:\/\/mathfun4kids.com\/mlog\/?p=3877","title":{"rendered":"Geometry Challenge – 15"},"content":{"rendered":"\n

In $\\triangle{ABC}$, $D$ is a point on $BC$. $\\angle{ABC}=100^\\circ$, $\\angle{BCA}=20^\\circ$, $\\angle{BAD}=50^\\circ$. Prove $AB=CD$.<\/p>\n\n\n\nClick here<\/a> for the proof.\n\n\n\n

\"\"<\/figure>\n\n\n\n
\n\n\n\n

Proof 1:<\/strong> Clearly, $\\angle{ADC}=150^\\circ$, and $\\angle{CAD}=10^\\circ$. According to the law of sines, we have:<\/p>\n\n\n\n

$ \\dfrac{sin\\angle{BCA}}{AB}=\\dfrac{sin\\angle{ABC}}{AC}$ and $ \\dfrac{sin\\angle{CAD}}{CD}=\\dfrac{sin\\angle{ADC}}{AC}$<\/p>\n\n\n\n

Therefore<\/p>\n\n\n\n

$AB=\\dfrac{sin\\angle{BCA}}{sin\\angle{ABC}}AC=\\dfrac{sin(20^\\circ)}{sin(100^\\circ)}AC=\\dfrac{2sin(10^\\circ)cos(10^\\circ)}{sin(80^\\circ)}AC=\\dfrac{2sin(10^\\circ)cos(10^\\circ)}{cos(10^\\circ)}AC$ <\/p>\n\n\n\n

$$=2sin(10^\\circ)AC$$<\/p>\n\n\n\n

$CD=\\dfrac{sin\\angle{CAD}}{sin\\angle{ADC}}AC=\\dfrac{sin(10^\\circ)}{sin(150^\\circ)}AC=\\dfrac{sin(10^\\circ)}{sin(30^\\circ)}AC=\\dfrac{sin(10^\\circ)}{\\dfrac{1}{2}}AC=2sin(10^\\circ)AC$<\/p>\n\n\n\n

We have $\\boxed{AB=CD}$.<\/p>\n\n\n\n

Proof 2<\/strong>: Clearly, $\\angle{BDA}=30^\\circ$, and $\\angle{BAC}=60^\\circ$. According to the law of sines, we have:<\/p>\n\n\n\n

$ \\dfrac{sin\\angle{BDA}}{AB}=\\dfrac{sin\\angle{BAD}}{BD}$ and $ \\dfrac{sin\\angle{BCA}}{AB}=\\dfrac{sin\\angle{BAC}}{BC}$<\/p>\n\n\n\n

Therefore<\/p>\n\n\n\n

$BD=\\dfrac{sin\\angle{BAD}}{sin\\angle{BDA}}AB=\\dfrac{sin(50^\\circ)}{sin(30^\\circ)}AB=\\dfrac{sin(50^\\circ)}{\\dfrac{1}{2}}AB=2cos(40^\\circ)AB$<\/p>\n\n\n\n

And<\/p>\n\n\n\n

$BC=\\dfrac{sin\\angle{BAC}}{sin\\angle{BCA}}AB=\\dfrac{sin(60^\\circ)}{sin(20^\\circ)}AB=\\dfrac{sin(60^\\circ+20^\\circ)}{sin(20^\\circ)}AB$<\/p>\n\n\n\n

$=\\dfrac{cos(40^\\circ)sin(20^\\circ)+sin(40^\\circ)cos(20^\\circ)}{sin(20^\\circ)}AB$<\/p>\n\n\n\n

$=\\dfrac{cos(40^\\circ)sin(20^\\circ)+2sin(20^\\circ)cos(20^\\circ)cos(20^\\circ)}{sin(20^\\circ)}AB$$<\/p>\n\n\n\n

$=(cos(40^\\circ)+2cos^2(20^\\circ))AB=(cos(40^\\circ)+cos(40^\\circ)+1)AB$<\/p>\n\n\n\n

$=(2cos(40^\\circ)+1)AB$<\/p>\n\n\n\n

Therefore $$CD=BC-BD=(2cos(40^\\circ)+1)AB-2cos(40^\\circ)AB=AB$$<\/p>\n\n\n\n

Proof 3:<\/strong> Let $E$ be on $AC$ so that $AB=AE$. <\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

As $\\angle{BAE}=60^\\circ$, $\\triangle{ABE}$ is equilateral. Therefore $\\angle{AEB}=60^\\circ$<\/p>\n\n\n\n

Because $\\angle{ADB}=30^\\circ$, $\\angle{AEB}=2\\angle{ADB}$, and $EA=EB$, $D$ must be on the circle with its center at $E$ and its radius as $EB$. Therefore $AE=BE=DE$. And $\\triangle{AED}$ is isosceles. <\/p>\n\n\n\n

We have $\\angle{CED}=\\angle{DAE}+\\angle{ADE}=10^\\circ+10^\\circ=20^\\circ$. Therefore $\\triangle{CDE}$ is isosceles. $CD=DE$. As $DE=BE=AE=AB$, we have $AB=CD$.<\/p>\n\n\n\n<\/div>\n","protected":false},"excerpt":{"rendered":"

In $\\triangle{ABC}$, $D$ is a point on $BC$. $\\angle{ABC}=100^\\circ$, $\\angle{BCA}=20^\\circ$, $\\angle{BAD}=50^\\circ$. Prove $AB=CD$. Click here for the proof. Proof 1: Clearly, $\\angle{ADC}=150^\\circ$, and $\\angle{CAD}=10^\\circ$. According to the law of sines, we have: $ \\dfrac{sin\\angle{BCA}}{AB}=\\dfrac{sin\\angle{ABC}}{AC}$ and $ \\dfrac{sin\\angle{CAD}}{CD}=\\dfrac{sin\\angle{ADC}}{AC}$ Therefore $AB=\\dfrac{sin\\angle{BCA}}{sin\\angle{ABC}}AC=\\dfrac{sin(20^\\circ)}{sin(100^\\circ)}AC=\\dfrac{2sin(10^\\circ)cos(10^\\circ)}{sin(80^\\circ)}AC=\\dfrac{2sin(10^\\circ)cos(10^\\circ)}{cos(10^\\circ)}AC$ $$=2sin(10^\\circ)AC$$ … Continue reading →<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"inline_featured_image":false},"categories":[9,15],"tags":[],"_links":{"self":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/3877"}],"collection":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=3877"}],"version-history":[{"count":51,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/3877\/revisions"}],"predecessor-version":[{"id":4614,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/3877\/revisions\/4614"}],"wp:attachment":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=3877"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=3877"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=3877"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}