# Prove that $B$, $C$, $M$, $N$ and $P$ are points of a cyclic polygon.

$$E$$, $$F$$ 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.

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.

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.

• How can you be sure that $T \in ED$? – Lê Thành Đạt Jul 20 '19 at 0:16
• @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. – sirous Jul 20 '19 at 7:53
• 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. – sirous Jul 20 '19 at 15:21
• It was a typo ,I corrected it. – sirous Jul 20 '19 at 16:09
• If $\widehat{TEC} = \widehat{BFT} = \widehat{FED} = \widehat{EDC}$ then $\widehat{TEC}$ should be equal to $\widehat{EDC}$, not $\widehat{DEC}$. – Lê Thành Đạt Jul 21 '19 at 5:53