3. Let $A B C D$ be a quadrilateral for which $\angle B A C=\angle A C B=20^{\circ}, \angle D C A=30^{\circ}$ and $\angle C A D=40^{\circ}$. Determine the measure of the angle $\angle C B D$.
III/3.

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1. method. We observe that triangle $ABC$ is isosceles with the vertex at $B$. Let $E$ be a point on line $CD$ such that $\angle CAE = 30^\circ$, so triangle $ACE$ is also isosceles with the vertex at $E$. Since $\angle CAD = 40^\circ$, point $E$ lies between $C$ and $D$. Then quadrilateral $ABCE$ is a kite, so its diagonals $AC$ and $BE$ intersect at a right angle. Therefore, $\angle BEC = 90^\circ - 30^\circ = 60^\circ$ and thus $\angle BAD = 60^\circ = \angle BEC = 180^\circ - \angle DEB$. This means that points $A, B, E,$ and $D$ are concyclic. By the inscribed angle theorem, $\angle EBD = \angle EAD = 40^\circ - 30^\circ = 10^\circ$, so $\angle CBD = \angle CBE + \angle EBD = (90^\circ - 20^\circ) + 10^\circ = 80^\circ$.

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2. method. We observe that triangle $ABC$ is isosceles with the vertex at $B$. Let $\mathcal{K}$ be the circle with center $B$ passing through points $A$ and $C$. We will prove that point $D$ also lies on circle $\mathcal{K}$. From the problem data, we calculate $\angle ADC = 180^\circ - 40^\circ - 30^\circ = 110^\circ$ and $\angle CBA = 180^\circ - 20^\circ - 20^\circ = 140^\circ$. Therefore, the central angle over arc $\widehat{AC}$ is $\angle ABC = 360^\circ - \angle CBA = 220^\circ$ and is twice the angle $\angle ADC$. By the theorem of the central and inscribed angles, $\angle ADC$ is an inscribed angle over arc $\widehat{AC}$, which means that point $D$ lies on circle $\mathcal{K}$. From this, it follows that triangle $CBD$ is isosceles with the vertex at $B$, so $\angle CBD = 180^\circ - 2 \angle DCB = 180^\circ - 100^\circ = 80^\circ$.

1. method:

Observation that $\triangle ABC$ is isosceles

1 point

Defining point $E$ such that $\angle CAE = 30^\circ$ 2 points

Observation that quadrilateral $ABED$ is concyclic 2 points

Calculation of $\angle CBD = 80^\circ$ 2 points

## 2. method:

Observation that $\triangle ABC$ is isosceles 1 point

Considering circle $\mathcal{K}$ with center at point $B$ passing through points $A$ and $C \ldots 2$ points

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![](https://cdn.mathpix.com/cropped/2024_06_07_77e56b77a012e45bd48fg-16.jpg?height=60&width=1633&top_left_y=2580&top_left_x=220)

## For a correct solution without a proper proof, 1 point is awarded.