Gonitzoggoβ
Editorial
Need a hint? Checkout the editorial.
View Editorial
Editorial
$L$ is the $C-$excenter of $\triangle CKM$, hence $KL$ is external bisector of $\angle CKM$, making $\angle CKM=60^\circ$
Mac Miller
Let $ABC$ be a triangle, and $K$ and $L$ be points on $AB$ such that $\angle ACK = \angle KCL = \angle LCB$. Let $M$ be a point in $BC$ such that $\angle MKC = \angle BKM$. If $ML$ is the angle bisector of $\angle KMB$, find $\angle MLC$.
Geometry
1 Upvote 0 Downvote
Need a hint? Checkout the editorial.
View Editorial
apudey0507