8,9
[Sum of angles in a triangle. Theorem about the exterior angle.]

On the sides $AB$ and $BC$ of triangle $ABC$, points $K$ and $L$ are chosen respectively, such that $\angle KCB = \angle LAB = \alpha$. Perpendiculars $BD$ and $BE$ are dropped from point $B$ to the lines $AL$ and $CK$ respectively. Point $F$ is the midpoint of side $AC$. Find the angles of triangle $DEF$.
Let $M$ and $N$ be the midpoints of sides $AB$ and $BC$ respectively. Denote $\angle B = \beta$. Then $\angle AMF = \angle CNF = \beta$. Since $EN$ and $DM$ are medians of the right triangles $CBE$ and $ADB$, drawn from the vertices of the right angles, we have $EN = \frac{1}{2} BC = NC = MF$,

$DM = \frac{1}{2} AB = AM = NF$.

Moreover, by the external angle theorem of a triangle, $\angle BNE = 2\alpha = \angle BMD$, so $\angle ENF = 180^\circ - \angle BNE - \angle CNF = 180^\circ - 2\alpha - \beta$,

$\angle DMF = 180^\circ - \angle BMD - \angle AMF = 180^\circ - 2\alpha - \beta$.

Therefore, triangles $ENF$ and $DMF$ are congruent by two sides and the included angle. Hence, $EF = DF$. Since $\angle MFD = \angle FEN$ and

$\angle NFE = \angle MDF$ (from the congruence of triangles $ENF$ and $DMF$), we have $\angle MFD + \angle NFE = \angle MFD + \angle MDF = 180^\circ - \angle DMF = 2\alpha + \beta$.

On the other hand, $\angle MFD + \angle NFE = \angle MFN + \angle DFE = \beta + \angle DFE$. From the equality $2\alpha + \beta = \beta + \angle DFE$, we find that $\angle DFE = 2\alpha$. Therefore,

$\angle DEF = \angle EDF = 90^\circ - \alpha$.

![](https://cdn.mathpix.com/cropped/2024_05_06_95d66af6ee6f6b4b6717g-14.jpg?height=864&width=826&top_left_y=1766&top_left_x=639)

## Answer

$2\alpha, 90^\circ - \alpha, 90^\circ - \alpha$.

In triangle $ABC$, the angle at vertex $A$ is $60^\circ$. Inside the triangle, a point $O$ is taken such that $\angle AOB = \angle AOC = 120^\circ$. Points $D$ and $E$ are the midpoints of sides $AB$ and $AC$. Prove that quadrilateral $ADOE$ is cyclic.

## Solution

Let $\angle OBA = \alpha$. Then $\angle BAO = 180^\circ - \angle AOB - \angle ABO = 60^\circ - \alpha$. Therefore, $\angle OAC = \angle BAC - \angle BAO = \alpha$. Hence, triangles $AOB$ and $COA$ are similar by two angles, from which it follows that $BD : AE = AB : AC = BO : AO$. Therefore, triangles $OBD$ and $OAE$ are similar by two sides and the included angle. Consequently, $\angle OEA = \angle ODB = 180^\circ - \angle ODA$, which is what we needed to prove.

![](https://cdn.mathpix.com/cropped/2024_05_06_95d66af6ee6f6b4b6717g-15.jpg?height=643&width=600&top_left_y=639&top_left_x=755)

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