In triangle $A B C$, side $A B$ is equal to 2, and angles $A$ and $B$ are equal to $60^{\circ}$ and $70^{\circ}$, respectively. On side $A C$, a point $D$ is taken such that $A D=1$.

Find the angles of triangle $B D C$.
Prove that triangle $ABD$ is a right triangle.

## Solution

$\angle C=180^{\circ}-\left(60^{\circ}+70^{\circ}\right)=50^{\circ}$.

On segment $DC$, take a point $M$ such that $DM=AD=1$. Then triangle $ABM$ is equilateral, and $BD$ is its median, and therefore, its altitude. Thus,

$\angle CBD=90^{\circ}-50^{\circ}=40^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_96d74c2697cd905bd948g-05.jpg?height=363&width=437&top_left_y=2071&top_left_x=822)

## Answer

$40^{\circ}, 90^{\circ}, 50^{\circ}$. Submit a comment

| The diagonals of quadrilateral $ABCD$, whose vertices lie on a circle, intersect at point $M$.

It is known that $\angle ABC=72^{\circ}, \angle BCD=102^{\circ}$,

$\angle AMD=110^{\circ}$. Find $\angle ACD$.

## Answer

$52^{\circ}$.

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