Geometry Problem. Post your solution in the comments box below.

Level: Mathematics Education, High School, Honors Geometry, College.

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## Friday, July 24, 2015

### Geometry Problem 1141: Inscribed Pentagon, Circle, Triangle, Congruence, Midpoint, Concurrency

Labels:
circle,
concurrent,
congruence,
inscribed,
midpoint,
pentagon

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Consider Tr.ABC , lets assume Angle ACB = x and Angle BAC =y ,

ReplyDeleteWe get Angle ABC = 180 - (x+y)

Tr. AFB and Tr. BMC are similar

AF/FB = BM/CM

Angle AFB = Angle BMC = 180 -x - y

Now consider Tr.AFG and Tr. NMC

AF/FG = MN/CM, angle AFG = angle NMC = x + y ,

Angle AGF = Angle MCN , by AA similarity Tr.AFG is similar to Tr. APC

Angle APC = Angle AFG = x + y

Quad. ABCP is cyclic hence P lies on circle O.

QED