A convex quadrilateral $ABCD$ has $\angle BAC = \angle ADC$. Let $M{}$ be the midpoint of the diagonal $AC$. The side $AD$ contains a point $E$ such that $ABME$ is a parallelogram. Let $N{}$ be the midpoint of the line segment $AE{}$. Prove that the line $AC$ touches the circumcircle of the triangle $DMN$ at point $M{}$.
Problem
Source: Estonia TST 2023
Tags: geometry
starchan
03.08.2023 13:25
nice
We've to show that $AM^2 = AN \cdot AD$. Let $X$ be the point such that $AXCB$ is a parallelogram and extend ray $CX$ to meet line $AD$ at point $Y$. Observe that $BX \parallel AY$ and $AB \parallel XY$ by definition, so $ABXY$ is a parallelogram. In particular, $E$ is the midpoint of $AY$. Therefore, proving $AM^2 = AN \cdot AD$ is equivalent to showing $AC^2 = 4AN \cdot AD = AD \cdot AY$ or that $\angle YCA = \angle CDA$. But this is clear, since $\angle CAD = \angle BAC = \angle XCA$ due to $BA \parallel XC$.
SQTHUSH
03.08.2023 14:51
Consider that $\angle BAM=\angle AME=\angle ADC$ Which means that $M,C,D,E$ are cyclic Note that $\angle AMD=\angle AEC=\angle ANM$ Hence $AM^{2}=AN\cdot AD$ Which means that $AM$ is tangent to $\odot(DMN)$
X.Allaberdiyev
11.08.2023 14:25
Too easy for TST. $\angle AME=\angle BAC=\angle ADC$, which means that $(M E D C)$ is cyclic. Then $AME$ and $ADC$ triangles are similar and $N$ is midpoint of $AE$, $M$ is midpoint of $AC$, which gives that $\angle AMN=\angle ADM$ (from similarity of triangles $AME$ and $ADC$), which completes the solution.
dkshield
02.12.2023 01:45
Let $\angle BAC = \angle ADC = \alpha$ Since $AB\parallel ME$ so $\angle BAC=\angle BAM=\alpha \angle AME$, since $\angle ADC$, so $MEDC$ is cyclic. Since $N$ is midpoint of $AE$ and $M$ is midpoint $AC$, so $\beta = \angle AMN=\angle ACE=\angle MDE=\angle AMN$ so $AC$ is tangent to $(MND)$, as desired.
torch
02.12.2023 04:32
By Power of a Point, we wish to prove $AN \cdot AD = AM^{2}$. Since $ABME$ is a parallelogram, $AE \| BM$. Thus, $\angle BAM = \angle AME$. We can now notice $MEDC$ is cyclic, as $\angle EMC + \angle DEC = 180^{\circ} - \angle AME +\angle DEC = 180^{\circ}$. So, by Power of a Point, $AM \cdot AC = AE \cdot AD$. Since $AC=2AM$, the equation can be rewritten as $AM^{2} = \frac{1}{2}\cdot AE \cdot AD$. Our desired result follows from $AN=\frac{1}{2} AE$.