Let $ABC$ with $AB<AC<BC$ be an acute angled triangle and $c$ its circumcircle. Let $D$ be the point diametrically opposite to $A$. Point $K$ is on $BD$ such that $KB=KC$. The circle $(K, KC)$ intersects $AC$ at point $E$. Prove that the circle $(BKE)$ is tangent to $c$.
Problem
Source: Greece JMO 2018, pr 4
Tags: geometry, tangent circles
enhanced
05.03.2018 23:06
Let $L_1,L_2$ be tangent to $\odot (ABC),\odot (BKE)$ at $B$ ,Now $\angle (L_1,BD)=90-\angle ACB$ Now since $KB=KE,\implies \angle KEB=\angle EBK$ also $\angle EKB=2\angle ACB,\implies$ $ \angle(L_2,BD) =\angle KEB=$ $90-\angle ACB $ so $\angle (L_1,BD)=\angle (L_2,BD),\implies L_1\equiv L_2$ so $\odot (ABC),\odot (BKE)$ are tangent to each other $\blacksquare$
khanhnx
28.01.2019 17:54
Through $B$ draw tangent $Bd$ with $c$ We have: $\widehat{dBE}$ = $\widehat{dBA}$ + $\widehat{ABE}$ = $\widehat{ACB}$ + $\widehat{ACB}$ = 2 $\widehat{ACB}$ = $\widehat{BKE}$ then $Bd$ tangent with ($BKE$)
parmenides51
30.04.2024 07:21
Let O be the circumcenter of (ABC), Lemma: <BOC=2<A => <OCB=<OBC=90^o-A BK=KC=KE=> BKE isosceles with bases BE Let the perpendicular bisector of BE, which is also an angle bisector of <BKE, intersect BO at point M. We shall prove that M is the circumcenter of (BKE) and since B-M-O (collinear) , we get that (BEK), (ABC) are tangent [as having their centers collinear with their touchpoint] <BME=1/2 <BKE=1/2 arc BE=2/2<ECB=<C <MBC=90^o-<A (lemma) <CMK=2 arc CD=<DAC=90^o-<B (lemma) So <MBK=<MBC+<CMK=180^o-<A-<B =<C Finally <MBK=<C=<BKM => BM =MK and since BM=Me (from the perp.bisector of EB) we get BM=MK=ME , which means that M is the circumcenter of (BEK)
redlovesmaths
26.05.2024 07:42
Another proof: Let F be on BC so that FB=FC. Angle chasing: BKE=2BCE ( K is the center of (BKE)) =2BCF=2(90-CFK)=180-BFE ==> BFEK is a cycli quadrilaterial. Now we shall prove ABF=CBE. AEB=FKB=90- CBK=90-CBD=90-CAD= ABC ==> AO is perpendicular to BE --> CBE=KOD=AOF=ABF (qed)