{"id":4686,"date":"2024-12-08T13:02:44","date_gmt":"2024-12-08T17:02:44","guid":{"rendered":"http:\/\/mathfun4kids.com\/mlog\/?p=4686"},"modified":"2025-02-24T14:06:20","modified_gmt":"2025-02-24T18:06:20","slug":"algebra-geometry-challenge-2","status":"publish","type":"post","link":"https:\/\/mathfun4kids.com\/mlog\/?p=4686","title":{"rendered":"Algebra\/Geometry Challenge \u2013 2"},"content":{"rendered":"\n

Cyclic quadrilateral $ABCD$ has lengths $BC=CD=3$, $AB=5$, $AD=8$. What is the length of the shorter diagonal of $ABCD$? Click here<\/a> for the solution.<\/p>\n\n\n\n

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Solution<\/strong>: Let $AC$ and $BD$ intersect at $E$, $x=BD$, $y=AE$, $z=CE$, $AC=y+z$. <\/a><\/p>\n\n\n\n

Because $BC=CD$, $AC$ is angle bisector of $\\angle{BAD}$. By Angle Bisector Theorem<\/a>, we have $\\dfrac{BE}{DE}=\\dfrac{AB}{AD}=\\dfrac{5}{8}$. Because $BE+DE=BD=x$, we have $BE=\\dfrac{5}{13}x$, $DE=\\dfrac{8}{13}x$.<\/p>\n\n\n\n

By Ptolemy’s Theorem<\/a>, we have $BD\\cdot AC=AB\\cdot CD+AD\\cdot BC$. Hence, $$x(y+z)=5\\cdot 3+3\\cdot 8=39\\tag{1}$$<\/p>\n\n\n\n

Because $BC=CD$, we have $\\angle{BAC}=\\angle{DAC}=\\angle{BDC}=\\angle{CAD}$.<\/p>\n\n\n\n

Because $\\angle{BCE}=\\angle{ADE}$, $\\triangle{BCE}\\sim\\triangle{ADE}$. Therefore $$\\dfrac{CE}{DE}=\\dfrac{BE}{AE}=\\dfrac{BC}{AD}\\ \\ \\ \\ i.e.\\ \\ \\ \\ \\dfrac{z}{\\dfrac{8}{13}}=\\dfrac{\\dfrac{5}{13}x}{y}=\\dfrac{3}{8}$$ We have $$y=\\dfrac{40}{39}x\\tag{2}$$ $$z=\\dfrac{3}{13}x\\tag{3}$$<\/p>\n\n\n\n

Solving equation (1), (2) and (3), and ignoring negative solutions, we have $$x=\\dfrac{39}{7}\\ \\ \\ \\ \\ \\ \\ y=\\dfrac{40}{7}\\ \\ \\ \\ \\ \\ z=\\dfrac{9}{7}$$<\/p>\n\n\n\n

Therefore, $BD=x=\\dfrac{39}{7}$, $AC=y+z=7$. Therefore, the length of the shorter diagonal is $\\boxed{\\dfrac{39}{7}}$.<\/p>\n\n\n\n

Note: since $\\triangle{BCE}\\sim\\triangle{ACB}$, we have $\\dfrac{BC}{CE}=\\dfrac{AC}{BC}$, i.e. $\\dfrac{3}{z}=\\dfrac{y+z}{3}$. We have $$z(y+z)=9\\tag{4}$$ We obtain the same result by solving equation (2), (3) and (4).<\/p>\n\n\n\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Cyclic quadrilateral $ABCD$ has lengths $BC=CD=3$, $AB=5$, $AD=8$. What is the length of the shorter diagonal of $ABCD$? Click here for the solution. Solution: Let $AC$ and $BD$ intersect at $E$, $x=BD$, $y=AE$, $z=CE$, $AC=y+z$. Because $BC=CD$, $AC$ is angle … 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":[13,9],"tags":[],"_links":{"self":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/4686"}],"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=4686"}],"version-history":[{"count":21,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/4686\/revisions"}],"predecessor-version":[{"id":4801,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/4686\/revisions\/4801"}],"wp:attachment":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=4686"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=4686"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=4686"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}