Thursday, July 16, 2015

Geometry Problem 1138: Square, Perpendicular, Angle, 90 Degrees, Triangle, Congruence

Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.

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Online Math: Geometry Problem 1138: Square, Perpendicular, Angle, 90 Degrees, Triangle, Congruence.

Posted by Antonio Gutierrez at 12:36 PM
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8 comments:

  1. AjitJuly 18, 2015 at 12:23 AM

    Join C to E & F and observe that triangles EBC & FDC are congruent. Hence, /_BEC=/_DFC -------(1).
    From Tr. EAD, we can say that DG*DE= AD²=DC². In other words, DC is tangent to the circum-circle of Tr.GEC and, therefore, we can see that /_GEC=/_GCD (Alternate segment theorem). Moreover, /_BEG=/_GDC since BE//DC. Thus, /_BEC=/_BEG+/_GEC= /_GDC+/_GCD= /_EGC. But by (1) above, /_BEC=/_DFC. This means that /_EGC = /_DFC implying that quad.GCFD is concyclic. Then, /_CGF=/_CDF=90°. QED.

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    Replies
    1. Sumith PeirisJuly 18, 2015 at 10:21 AM

      Got almost all your steps Ajit without realising that DC is tangential to GEC Tr. Hence could not complete

      Well done

      Delete
    2. AjitJuly 18, 2015 at 11:03 PM

      That, in fact, is what took me almost two hours to realize. Otherwise, I was planning to do this by analytical geometry in this fashion: Let A:(0,0), D:(a,0),F:(a+p,0), C:(a,a) and B:(0,a+p). ED can be written as: x/a+y/(a+p)=1 while AG is: y=ax/(a+p) which gives us
      G as :[a(a+p)²/(a²+(a+p)²), a²(a+p)/(a²+(a+p)²)]. Now we determine slope of CG as [(a²+p²+ap) /a²] and that of GF as [-a²/(a²+p²+ap) ] , the product of which is -1. Hence etc.
      But this solution misses all the niceties, the congruence of two triangles, the tangency and the consequent concyclicity etc of the solution by Euclidean elements.

      Delete
    3. Benjamin LeisMay 4, 2018 at 2:31 PM

      Just leaving a breadcrumb for . The crucial tangency of DC and Tr. GEC comes from the power of a point theorem.

      Delete
  2. Stan FulgerAugust 6, 2015 at 12:35 AM

    E, F are symmetrical about AC, so if H is intersection of EF with AC, EF_|_AC, and EG.ED=AE^2=EH.EF, so GDFH is cyclic, as well as DFCH, hence CH_|_GF.

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  3. piet van KampenNovember 24, 2015 at 11:55 AM

    Extend AG and BC ; intersect at P.
    tr. APB = tr. DEA ( AE=BP => CP=DF
    DFPC is parallellogram ( CP//DEF and CP=DF ) ; DFPC is rectangle thus concyclic
    GDCP is concyclic
    DPFC and GDCP concyclic => GDFCP concyclic => <CGF=<CDF=90

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  4. Sumith PeirisOctober 26, 2018 at 2:35 AM

    If EF and AC cut at X then
    AXE, AXF and AEF are all 45-45-90 Tr.s

    So AEXG is concyclic and
    < XGC = < AEX = 45 = < EAX = EGX = < EFD

    Hence XGDF is concyclic and so
    < CGF = CXF = 90

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  5. Stan FULGERJuly 1, 2021 at 11:16 AM

    My new solution at https://stanfulger.blogspot.com/2021/07/problem-1138-gogeometry.html

    ReplyDelete

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