An acute triangle is a triangle that has all angles less that $90^{\circ}$ ($90^{\circ}$ is a Right Angle). Let $ABC$ be a right-angled triangle with $\angle ACB =90^{\circ}.$ Let $CD$ be the altitude from $C$ to $AB,$ and let $E$ be the intersection of the angle bisector of $\angle ACD$ with $AD.$ Let $EF$ be the altitude from $E$ to $BC.$ Prove that the circumcircle of $BEF$ passes through the midpoint of $CE.$
Problem
Source: Canada Junior MO 2023 P2
Tags: geometry, angle bisector, circumcircle
PEKKA
19.01.2024 19:20
Let $M$ denote the midpoint of $CE.$ Now we want to show that $B,E,M,F$ are concyclic. Let $\angle ABC=2 \alpha$ and let $\angle CAB=2 \beta.$ Then angle chasing tells us that $\angle MFB+\angle MEB=180$ and we're done
AshAuktober
20.01.2024 08:37
Let the midpoint of $CE$ be $X$. Denote by $\angle A$ the angle $\angle CAB$. Claim 1: $BE = BC$. Proof: Note that $\angle BCE = \angle DCE + \angle DCB = \frac{1}{2} \angle ACD + \angle CAD = 45^\circ - \frac{\angle A}{2} + \angle A = 45^\circ + \frac{\angle A}{2}$. Also see that $\angle BCE = 180^\circ - \angle BCE - \angle EBC = 180^\circ - (45^\circ + \frac{\angle A}{2}) - (90^\circ - \angle A) = 45^\circ + \frac{\angle A}{2}$. This gives us $\angle BCE = \angle BEC \implies BE = BC.$ $\square$ Note that Claim 1 implies that $BX \perp CE$, i.e. $\angle BXE = 90^\circ$. However since $\angle BFE = 90^\circ$, we have that $BFXE$ is cyclic. $\square$
Attachments:
parmenides51
08.05.2024 14:36
Let $M$ be the midpoint of $CE$ and $\angle ACD=2a$
$\angle ACE=\angle ECD=a$,
$\angle CED=90^o -a$ (angles sum in right $CED$) (1)
$\angle DCB=90^o- 2a$
$\angle ECB=\angle ECD +\angle DCB= a+ 90^o -2a=90^o -a =\angle CED$ because of (1)
so $ECB$ is isosceles with base $CE \Rightarrow \angle EMB=90^o$
so $\angle EFB=90^o=\angle EMB=90^o \Rightarrow EMFB$ cyclic
midpoint of $CE$ is third altitude in triangle $BCE$ as $BCE$ is isosceles ($BC=BE$)