Pentagon $ABCDE$ with $CD\parallel BE$ is inscribed in circle $\omega$. Tangent to $\omega$ through $B$ intersects line $AC$ at $F$ in a way that $A$ lies between $C$ and $F$. Lines $BD$ and $AE$ intersect at $G$. Prove that $FG$ is tangent to the circumcircle of $\triangle ADG$.
Problem
Source: Latvia BW TST 2021 P9
Tags: geometry, circumcircle, proposedbyQaqash
kent2207
27.09.2021 05:45
Please check my solution We have $\angle$$FAG$$=\angle$$EAC$$=\angle$$EBC$$=\angle$$DEB$ (because $BE$ is parallel to $DC$) $=\angle$$FBG$ ($FB$ is a tangent to $\omega$) So $F,A,B,G$ are cyclic Then we have $\angle$$FGA$$=\angle$$FBA$$=\angle$$ADG$ So $FG$ is a tangent to $(ADG)$ (Q.E.D)
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Blastoor
27.09.2021 11:47
kent2207 wrote: Please check my solution We have $\angle$$FAG$$=\angle$$EAC$$=\angle$$EBC$$=\angle$$DEB$ (because $BE$ is parallel to $DC$) $=\angle$$FBG$ ($FB$ is a tangent to $\omega$) So $F,A,B,G$ are cyclic Then we have $\angle$$FGA$$=\angle$$FBA$$=\angle$$ADG$ So $FG$ is a tangent to $(ADG)$ (Q.E.D) The solution is correct.
kent2207
27.09.2021 18:53
Blastoor wrote: kent2207 wrote: Please check my solution We have $\angle$$FAG$$=\angle$$EAC$$=\angle$$EBC$$=\angle$$DEB$ (because $BE$ is parallel to $DC$) $=\angle$$FBG$ ($FB$ is a tangent to $\omega$) So $F,A,B,G$ are cyclic Then we have $\angle$$FGA$$=\angle$$FBA$$=\angle$$ADG$ So $FG$ is a tangent to $(ADG)$ (Q.E.D) The solution is correct. Thank you for your checking
TheCoolDinosuar
06.03.2023 11:05
By pascal on $BBEACD$, we get $FG\parallel CD$. Then \[\measuredangle FGD=\measuredangle CDB=\measuredangle EAB=\measuredangle GAD.\]Then $FG$ is tangent to $(AFG)$.