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.

5 comments:

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

      Well done

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    2. 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.

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  2. 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. 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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