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$E$, $F$ and $D$ are points respectively on line $CA$, $AB$ and $BC$ such that $CE = EA$, $AF = FB$ and $ED = DF$. Given point $P$ sastified $\widehat{CET} = 90^\circ - \dfrac{\widehat{EDF}}{2} = \widehat{TFB}$. The tangent of the circumcircle of $\triangle{TEF}$ at point $T$ intersects line $CA$ and $AB$ respectively at points $M$ and $N$. The circumcircle of $\triangle{AFE}$ intersects line $AT$ at point $T$ $(P \not\equiv A)$. Prove that $B$, $C$, $M$, $N$ and $P$ are points of a cyclic polygon.

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This might come as the most difficult geometry problem I've ever encountered. There's a clue and it is to let $Q = AP \cap (T, E, F)$ and prove that $AT$ and $AD$ are reflected one another in the bisector of $\widehat{CAB}$. $(AQ < AP)$

Working around the clue, I still can't figure it out. I suppose more points need to be set up.

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Draw the height of triangle FDE from vertex D. Since ED=DF this height is the bisector of $\angle FDE$. Hence we have:

$\angle DEF=\angle DFE=90-\frac{\angle EDF}{2}=\angle CET=\angle TFB$

FB is parallel to BC therefore:

$\angle EDC=\angle DEF=\angle CET$

Also:

$\angle AEF+\angle DEF+\angle DEC=180^o$

$\angle ACB=\angle AEF$

$\angle DEF=\angle EDC$

$\angle AEF+\angle FED+\angle DEC=180^o$

Due to data in question:

$\angle TEC=\angle BFT=\angle FED=\angle EDC$

$\angle TEC=\angle DEC$

That is T locates on ED and triangle EDC is isosceles,Also triangle FNT is isosceles because NT and NF are tangent to circumcircle of $\triangle TEF$ so we have:

$\angle ECD=\angle EDF=\angle FNT$

Therefore we have :

$\angle BCM=\angle BNM$

This is possible only if points B, C, M and M are on a circle, because the sides of angles BNM and BCM cross each other.Similar reasoning can be used for point P.

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  • $\begingroup$ How can you be sure that $T \in ED$? $\endgroup$ – Lê Thành Đạt Jul 20 '19 at 0:16
  • $\begingroup$ @LêThànhĐạt, The reason is:$\angle AEF+\angle DEF+\angle DEC=180^o$ $\angle ACB=\angle AEF$ $\angle DEF=\angle EDC$ $\angle AEF+\angle AEF+\angle AEF=180^o$ Due to data in question: $\angle TEC=\angle BET=\angle FED=\angle EDC$ ⇒$\angle TEC=\angle DEC$ That is T locates on ED . I edited my answer. $\endgroup$ – sirous Jul 20 '19 at 7:53
  • $\begingroup$ I never said $3,AEF=180^o$, $AEF=ACB ≠EFD$ and the sum of internal angles of triangle DEC is $180^o$ which is equal to sum of angles <AEF=<ACB, <EDC and <DEC. $\endgroup$ – sirous Jul 20 '19 at 15:21
  • $\begingroup$ It was a typo ,I corrected it. $\endgroup$ – sirous Jul 20 '19 at 16:09
  • $\begingroup$ If $\widehat{TEC} = \widehat{BFT} = \widehat{FED} = \widehat{EDC}$ then $\widehat{TEC}$ should be equal to $\widehat{EDC}$, not $\widehat{DEC}$. $\endgroup$ – Lê Thành Đạt Jul 21 '19 at 5:53

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