Monday, September 21, 2009

Problem 358. Isosceles triangle 80-80-20, Circle, Angles, Congruence

Proposed Problem
Click the figure below to see the complete problem 358 about Isosceles triangle 80-80-20, Circle, Angles, Congruence.

 Problem 358. Isosceles triangle 80-80-20, Circle, Angles, Congruence.
See more:
Complete Problem 358
Level: High School, SAT Prep, College geometry

5 comments:

  1. This comment has been removed by a blog administrator.

    ReplyDelete
  2. http://ahmetelmas-geo-geo-antonio.blogspot.com/

    ReplyDelete
  3. Let AD cut the circle at P and EF at Q.

    Easily Tr. AEC is 80-20-80 and so Tr. CEF is equilateral. Hence < DAF = 1/2 *60 - 20 = 10 and similarly < EFA = 1/2 *< ECA = 10. So AEPF is an isoceles trapezoid with AE = PF

    Now since < ACF = 80, < PFD = 140 and since AD = BD, < PDF = DPF = 40. Hence PF = DF and so AE = DF

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  4. Problem 358
    Triangle ECF is equilateral. The EF intersects AC in P (P is left of A ).Then <PEA=<EPA=40 so PA=AE.But triangle ACD= triangle FCP (AC=CF, <CPF=40=ADC,<PCF=<ACD=80). Therefore
    CD=PC or CF+FD=CA+AP so DF=AP=AE.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

    ReplyDelete
  5. Rotate the triangle ACD (60-80-40) clockwise over C by 20 degrees such that AC overlaps EC. Denote the new position of D as P (E is the new position of A). Let CP intersect the pink circle at Q.
    We can see that ECF is an equilateral triangle and AEC is an 80-80-20 triangle.
    Since m(ECP)=m(ACD)=80 and m(ECF)=60=> m(DCP)=20 and as DC=CP per construction => DCP is a 80-20-80 isosceles triangle
    As CF=CQ=Radius of the circle => DF=PQ and FQ//DP. On angle chasing we can see that FPQ is a 40-100-40 isosceles and FQ=PQ
    Since FCQ is an 80-20-80 isosceles and FC=AC=CQ=EC => FCQ is congruent to ACE
    => AE=FQ=PQ=DF

    ReplyDelete