See the equivalence with http://www.mathlinks.ro/Forum/viewtopic.php?t=126138 ? darij

Points $ A$, $ X$, $ A_{1}$,$ A_{2}$,$ B_{1}$,$ B_{2}$,$ C_{1}$,$ C_{2}$ are cyclic. From trigonometric form of Ceva theorem it is sufficiantly to prove that $ \frac{B_{1}A_{2}}{A_{2}C_{1}}\frac{C_{1}B_{2}}{B_{2}A_{1}}\frac{A_{1}C_{2}}{C_{2}B_{1}}=1$ Let $ D$, $ E$, $ F$ are feet of the altitudes from vertices $ A$, $ B$, $ C$ respectively. Points $ A$, $ H$, $ A_{2}$, $ E$ are cyclic. $ \angle B_{1}HA_{2}=\angle EHA_{2}$ ( or $ \angle B_{1}HA_{2}=180-\angle EHA_{2}$ ) $ \angle EHA_{2}=\angle EAA_{2}$ ( or $ \angle EHA_{2}=180-\angle EAA_{2}$ ) $ \angle EAA_{2}=\angle CAX$ ( or $ \angle EAA_{2}=180-\angle CAX$ ) $ sin\angle B_{1}HA_{2}=sin\angle CAX$ The same conclusion we can get using directed angles. Similarly $ sin\angle C_{1}HA_{2}=sin\angle BAX$, $ sin\angle C_{1}HB_{2}=sin\angle ABX$, $ sin\angle A_{1}HB_{2}=sin\angle CBX$, $ sin\angle A_{1}HC_{2}=sin\angle BCX$, $ sin\angle B_{1}HC_{2}=sin\angle ACX$ But from trigonometric form of Ceva theorem $ \frac{sin\angle CAX}{sin\angle BAX}\frac{sin\angle ABX}{sin\angle CBX}\frac{sin\angle BCX}{sin\angle ACX}=1$

Just for sake of completeness, note that this problem has been posted in the same form here: http://www.mathlinks.ro/viewtopic.php?t=80025.

Shishkin wrote: Points $ A$, $ X$, $ A_{1}$,$ A_{2}$,$ B_{1}$,$ B_{2}$,$ C_{1}$,$ C_{2}$ are cyclic. From trigonometric form of Ceva theorem it is sufficiantly to prove that $ \frac {B_{1}A_{2}}{A_{2}C_{1}}\frac {C_{1}B_{2}}{B_{2}A_{1}}\frac {A_{1}C_{2}}{C_{2}B_{1}} = 1$ i dun understand..why Ceva can be used in this way? I can't find out which triangle Ceva theorem is applied to it...and why sin C_1HB_2=sin ABX?? plz help.

Shishkin wrote: Points $ A$, $ X$, $ A_{1}$,$ A_{2}$,$ B_{1}$,$ B_{2}$,$ C_{1}$,$ C_{2}$ are cyclic. From trigonometric form of Ceva theorem it is sufficiantly to prove that $ \frac {B_{1}A_{2}}{A_{2}C_{1}}\frac {C_{1}B_{2}}{B_{2}A_{1}}\frac {A_{1}C_{2}}{C_{2}B_{1}} = 1$ yeah , I , don't understand too

1)the triangle$ABC$and$A_1B_1C_1$are similar. 2)$A_2A_1B_1=XAC$and similar othe angle:) 3)because that $XA,XB,XC$are concurrent in triangle $ABC$ than $A_1A_2,B_1B_2,C_1C_2$are concurrent in traingle $A_1B_1C_1$:) im so lazy to write compleat solution if have question ask it

My Solution : Let feet of perpendiculars of $A,B,C$ be respectively $A',B',C'$ We've $\angle {B_1A_1C_1}=\angle {B'HC}=\angle {C'HB}=A$ and same for $B$ and $C$. Which implies $A_1B_1C_1$ is similar to $ABC$. Now $\angle {B_1A_1A_2}=\angle{B_1HA_2}=\angle {XAC}$ same for $B,C$ also. So $A_1A_2$ correspondence to the Isogonal of $AX$. Same for $B,C$. Now since $AX,BX,CX$ are concurrent hence their Isogonals are also concurrent. So done