computational inside a 80-80-20 triangle, BC=12, ABE= 30^o , EF = FC

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Apr 2, 2025

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Mathdreams 4

N 14 minutes ago by Tony_stark0094

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Leo890 9

N an hour ago by jordiejoh

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Let $ABC$ be a triangle and $I$ its incenter. The lines $BI$ and $CI$ intersect the circumcircle of $ABC$ again at $M$ and $N$ , respectively. Let $C_1$ and $C_2$ be the circumferences of diameters $NI$ and $MI$ , respectively. The circle $C_1$ intersects $AB$ at $P$ and $Q$ , and the circle $C_2$ intersects $AC$ at $R$ and $S$ . Show that $P$ , $Q$ , $R$ and $S$ are concyclic.

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jordiejoh

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computational inside a 80-80-20 triangle, BC=12, ABE= 30^o , EF = FC

parmenides51 1

N Dec 21, 2020 by LeonhardEuler0

Source: December 2020 Ukraine Geometry Olympiad VIII p4

In an isosceles triangle $ABC$ with an angle $\angle A= 20^o$ and base $BC=12$ point $E$ on the side $AC$ is chosen such that $\angle ABE= 30^o$ , and point $F$ on the side $AB$ such that $EF = FC$ . Find the length of $FC$ .

1 reply

parmenides51

Dec 21, 2020

LeonhardEuler0

Dec 21, 2020

computational inside a 80-80-20 triangle, BC=12, ABE= 30^o , EF = FC

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Source: December 2020 Ukraine Geometry Olympiad VIII p4

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parmenides51

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parmenides51

#1 h Dec 21, 2020, 8:58 AM

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LeonhardEuler0

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LeonhardEuler0

#2 h Dec 21, 2020, 1:53 PM

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If you do some angle chasings Its easy to see:
$\angle EBC=\angle BEC=50^o$ so,
$BC=EC=12$ ,let $\angle FEB=a$ , Then you reflect the point $E$ according to $AB$ ,to create an equilateral triangle $(BEE')$ .Draw a line $E'C$ so $FE'C$ is isosceles triangle.( $\angle E'FC=140^o$ )
Then $(E'BC)$ and $(E'EC)$ are congruent triangles.Thus, $20+a=40-a$ so, $a=10^o$ .
So, $(FEC)$ is an equilateral triangle.
Henceforth, $FE=FC=EC=BC=12$

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