Let $M$ be the midpoint of cathetus $AB$ of triangle $ABC$ with right angle $A$. Point $D$ lies on the median $AN$ of triangle $AMC$ in such a way that the angles $ACD$ and $BCM$ are equal. Prove that the angle $DBC$ is also equal to these angles.
Problem
Source: Sharygin Finals 2023 10.1
Tags: geometry, Sharygin Geometry Olympiad, Sharygin 2023
X.Allaberdiyev
02.08.2023 19:00
Let $K$ be the midpoint of $BC$, then $MK//AC$ then we have $\angle NAC=\angle NCA=\angle NMK$ and $\angle KCM=\angle DCA$, which gives us that the triangles $ACD$ and $MCK$ are similar. Then $DC/CK=AC/MC$, then $DC/(2CK)=(AC/2)/MC$, then $DC/BC=MK/MC$ (Because $MK$ is midline) $(1)$. $\angle KMC=\angle MCA=\angle BDC$ $(2)$. Using $(1)$ and $(2)$ we obtain that $KMC$ and $DBC$ triangles are similar, then $\angle DBC=\angle MCK=\angle MCB$, which completes the solution.
VicKmath7
03.08.2023 19:40
Redefine $D$ as the foot of the perpendicular from the midpoint $O$ of $BC$ to the $C$-symmedian. Since $\angle AOD=\angle DAB$ due to $D \in (AOB)$ (well-known theory for humpty-dumpty points) and $\angle ADO=\angle BAC=90^{o}$, we have that $\angle DAC=\angle DCO=\angle ACN=\angle CAN$, so $D \in AN$ and the new definition is correct. Thus, $\angle DBC=\angle DAO=\angle CAO-\angle CAD=\angle ACB-\angle ACN=\angle BCM$, as desired.
NO_SQUARES
16.05.2024 08:37
Let $K$ be point such a $AKBC$ is parallelogram. So, $C, M, K$ are collinear. Since $\angle A = \pi/2$, we have $\angle DAC = \angle ACN = \angle BKC$. Also $\angle ACD=\angle BCM =\angle AKC$. We have $\Delta KBC \sim \Delta ADC$. So $CB/CD=CK/CA \Rightarrow \Delta CBD \sim CKA \Rightarrow \angle CBD = \angle CKA = \angle ACD$. $\blacksquare$