Let $ABC$ be a non-right isosceles triangle such that $AC = BC$. Let $D$ be such a point on the perpendicular bisector of $AB$, that $AD$ is tangent on the $ABC$ circumcircle. Let $E$ be such a point on $AB$, that $CE$ and $AD$ are perpendicular and let $F$ be the second intersection of line $AC$ and the circle $CDE$. Prove that $DF$ and $AB$ are parallel.
Problem
Source: 2019 Slovenia 1st TST P1
Tags: TST, geometry
MessingWithMath
19.02.2019 01:51
Solution: Denote K as projection of C onto AD. Suffices to prove <FDC=90 or CF diameter of circle ECD. This is equivalent to proving AC passes through center of this circle, or suffices to prove that AC is perpendicular bisector of ED. This means it suffices to prove triangle CEA is congruent to triangle CDA. We know that CA is shared side and <CEA = <CDA as <CEA=90-<KAE=90-<DAB=<CDA. Thus, suffices to prove <ECA = <DCA. Since <DCA = 1/2 <BCA, suffices to prove that <ECD = <ACB. Yet, since <ECD = <AOD (O circumcenter of ABC) as OA is perpendicular to AD so OA//KC=EC, and <AOD = 1/2 <AOB = 1/2 (2 <ACB) = <ACB. So we're done. Note how B is also on (ECD) which is quite cool!
Leartia
19.02.2019 03:54
This is way too easy for Slovenia It's straightforward to prove that $\triangle CAE \cong \triangle CAD.$ Let $CD \cap (ABC)=L$ $(L \neq C)$ $C$ is the center of a spiral symmetry that sends $FD$ to $LB.$ Since $\angle LBA = \angle LCB$ we get the desired result. Remark: After the congruence, one can simply use the fact that $A$ is, by definition, the orthocenter of $\triangle CED$ and finish off by reflecting the orthocenter lemma.
Mahdi_Mashayekhi
01.01.2022 14:07
Let CD meet AB at S and AD meet CE at K. we have ∠AKE = 90 = ∠ASD so DSKE is cyclic. ∠CEA = ∠ADS = ∠SDB ---> CEDB is cyclic. we know A is orthocenter of CDE so F is reflection of A across DE and AD = FD. ∠AFD = ∠FAD = ∠KAC = ∠ABC = ∠BAC ---> BA || DF.