{"id":3023,"date":"2022-11-18T04:36:53","date_gmt":"2022-11-18T08:36:53","guid":{"rendered":"http:\/\/mathfun4kids.com\/mlog\/?p=3023"},"modified":"2024-10-25T12:13:47","modified_gmt":"2024-10-25T16:13:47","slug":"2022-amc-10b-problem-20","status":"publish","type":"post","link":"https:\/\/mathfun4kids.com\/mlog\/?p=3023","title":{"rendered":"2022 AMC 10B Problem 20"},"content":{"rendered":"\n
\"\"<\/figure><\/div>\n

Let $ABCD$ be a rhombus with $\\angle{ADC}=46^{\\circ}$, and let $E$ be the midpoint of $\\overline{CD}$, and let $F$ be the point on $\\overline{BE}$ such that $\\overline{AF}$ is perpendicular to $\\overline{BE}$. What is the degree measure of $\\angle{BFC}$?<\/p>\n\n\n\n

(A)<\/strong> 110      (B) <\/strong>111      (C)<\/strong> 112      (D)<\/strong> 113      (E) <\/strong>114      Solution<\/a><\/p>\n\n\n\n

\n\n\n\n

Draw the diagonals of rhombus $ABCD$, $\\overline{AC}$ and $\\overline{BD}$, with $\\overline{AC}$ intersecting $\\overline{BD}$ at $G$, $\\overline{BD}$ intersecting $\\overline{AF}$ at $H$, and $\\overline{AC}$ intersecting $\\overline{BE}$ at $I$. Then draw line $\\overline{FG}$, and $\\overline{EG}$. We have the following diagram:<\/p>\n\n\n\n

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

Proposition 1: <\/strong>$ABFG$ is cyclic quadrilateral<\/a>, because both $\\triangle{ABG}$ and $\\triangle{ABF}$ are right triangles, as $\\overline{AG}\\perp\\overline{BG}$ and $\\overline{AF}\\perp\\overline{BF}$. Therefore $$\\angle{AFG}=\\angle{ABG}=\\angle{ADG}=\\dfrac{1}{2}\\angle{ADC}=23^{\\circ},\\ \\angle{FAG}=\\angle{FBG}$$ $$\\angle{GFI}=\\angle{HFI}-\\angle{HFG}=90^{\\circ}-23^{\\circ}=67^{\\circ}$$<\/p>\n\n\n\n

Proposition 2:<\/strong> $CEGF$ is cyclic quadrilateral<\/a>, because $$\\angle{GFE}=\\angle{GFI}=67^{\\circ}=90^{\\circ}-23^{\\circ}=\\angle{CGD}-\\angle{CDG}=\\angle{GCD}=\\angle{GCE}$$ and $$\\angle{FGC}=\\angle{AFG}+\\angle{FAG}=23^{\\circ}+\\angle{FBG}=\\angle{BDE}+\\angle{EBD}=\\angle{BEC}=\\angle{FEC}$$<\/p>\n\n\n\n

Because $E$ is midpoint of $CD$ and $\\triangle{CDG}$ is a right triangle, we have $$\\overline{CE}=\\overline{GE}=\\overline{ED}$$ Therefore $\\triangle{CEG}$ is an isosceles triangle, with $\\angle{CGE}=\\angle{GCE}=67^{\\circ}$. <\/p>\n\n\n\n

Because $CEGF$ is cyclic quadrilateral<\/a>, we have $\\angle{CFE}=\\angle{CGE}=67^{\\circ}$. Therefore $\\angle{BFC}=180^{\\circ}-\\angle{CFE}=180^{\\circ}-67^{\\circ}=113^{\\circ}$. So the answer is $\\boxed{(D)}$.<\/p>\n\n\n\n

Note: <\/em>We can also approve $CEGF$ is cyclic as $\\triangle{FIG}\\sim\\triangle{CIE}$, and $\\dfrac{\\overline{FI}}{\\overline{GI}}=\\dfrac{\\overline{CI}}{\\overline{EI}}$, i.e. $\\overline{CI}\\cdot\\overline{GI}=\\overline{EI}\\cdot\\overline{EF}$.<\/p>\n\n\n\n<\/div>\n\n\n\n

<\/p>\n","protected":false},"excerpt":{"rendered":"

Let $ABCD$ be a rhombus with $\\angle{ADC}=46^{\\circ}$, and let $E$ be the midpoint of $\\overline{CD}$, and let $F$ be the point on $\\overline{BE}$ such that $\\overline{AF}$ is perpendicular to $\\overline{BE}$. What is the degree measure of $\\angle{BFC}$? (A) 110      (B) 111      (C) 112      (D) 113      (E) 114      Solution Draw the … 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],"tags":[],"_links":{"self":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/3023"}],"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=3023"}],"version-history":[{"count":37,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/3023\/revisions"}],"predecessor-version":[{"id":3065,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/3023\/revisions\/3065"}],"wp:attachment":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=3023"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=3023"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=3023"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}