[ The angle between two chords and two secants ] [ The inscribed angle is half of the central angle ]

Chords $A C$ and $B D$ are drawn from the ends of arc $A B$, which contains $m^{\circ}$, such that the angle $D M C$, formed by their intersection, is equal to the inscribed angle $D N C$ in arc $C D$. Find the degree measure of this arc.
$D M C=\frac{U A B+U D N C}{2}$.

## Solution

Let the degree measure of arc $D N C$ be $n^{\circ}$. Then

$$
\angle D N C=\frac{\cup D A C}{2}=\frac{360^{\circ}-n^{\circ}}{2}=180^{\circ}-\frac{n^{\circ}}{2}
$$

Since $\angle D M C=\frac{\cup A B+U D N C}{2}$, then

$$
180^{\circ}-\frac{n^{\circ}}{2}=\frac{m^{\circ}+n^{\circ}}{2}
$$

From this, we find that $n^{\circ}=180^{\circ}-\frac{m^{\circ}}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_93426a107eaf3f76289cg-31.jpg?height=458&width=458&top_left_y=2215&top_left_x=800)

Answer

$180^{\circ}-\frac{m^{\circ}}{2}$.

In quadrilateral $A B C D$, angles $B$ and $D$ are right angles. Diagonal $A C$ forms an acute angle of $40^{\circ}$ with side $A B$, and an angle of $30^{\circ}$ with side $A D$. Find the acute angle between diagonals $A C$ and $B D$.

## Hint

The angle between intersecting chords is equal to half the sum of the opposite arcs cut off on the circle by these chords.

## Solution

Points $A, B, C$, and $D$ lie on a circle with diameter $A C$. Therefore,

$$
\begin{gathered}
\cup A B=2 \angle A C B=2\left(90^{\circ}-\angle C A B\right)=2\left(90^{\circ}-40^{\circ}\right)=100^{\circ} \\
\cup D C=2 \angle D A C=60^{\circ}
\end{gathered}
$$

If $O$ is the point of intersection of the diagonals, then

$$
\angle A O B=\frac{\cup A B+\cup D C}{2}=\frac{160^{\circ}}{2}=80^{\circ}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_93426a107eaf3f76289cg-32.jpg?height=443&width=504&top_left_y=1128&top_left_x=777)

## Answer

$80^{\circ}$.