Proposed Problem
Click the figure below to see the complete problem 358 about Isosceles triangle 80-80-20, Circle, Angles, Congruence.
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Complete Problem 358
Level: High School, SAT Prep, College geometry
Monday, September 21, 2009
Problem 358. Isosceles triangle 80-80-20, Circle, Angles, Congruence
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ReplyDeletehttp://ahmetelmas-geo-geo-antonio.blogspot.com/
ReplyDeleteLet AD cut the circle at P and EF at Q.
ReplyDeleteEasily 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
Problem 358
ReplyDeleteTriangle 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
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.
ReplyDeleteWe 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