In quadrilateral $A B C D A B=B C=m, \angle A B C=\angle A D C=120^{\circ}$. Find $B D$.
Consider a circle with center $B$ and radius $m$, which passes through points $A$ and $C$ (see figure).

![](https://cdn.mathpix.com/cropped/2024_05_06_8c5c78c6462ff090b828g-05.jpg?height=388&width=412&top_left_y=2350&top_left_x=823)

From the equality of angles $A B C$ and $A D C$, it follows that points $B$ and $D$ lie in different half-planes relative to the line $AC$. Moreover,

$\angle A D C=180^{\circ}-1 / 2 \angle A B C$, so point $D$ also lies on this circle. Therefore, $B D=m$.

Complexity: $3+$ Classes: $9,10,11$

[ $\left.\begin{array}{ll}\text { Auxiliary circle }\end{array}\right]$
![](https://cdn.mathpix.com/cropped/2024_05_06_8c5c78c6462ff090b828g-05.jpg?height=2120&width=524&top_left_y=0&top_left_x=1554)
[Pythagorean Theorem (direct and inverse) $)$ ]

Complexity: $3+$
Classes: $8,9,10,11$

## [ Inscribed and circumscribed circles ] Problem 66703 Topics: [ Inscribed quadrilaterals (other).] [ Median to the hypotenuse ]

Complexity: $3+$

Classes: $8,9,10,11$

Author: Bakayev E.V.

Point $\$ 0 \$$ is the center of the circumscribed circle of the acute-angled triangle $\$ \mathrm{ABC} \$$, \$ the foot of the perpendicular dropped from point \$ through the midpoint of side $\$ \mathrm{AB} \$$.

## Solution

\$ \angle $\mathrm{B}=$ \frac $\{1\}\{2\}$ \angle $\mathrm{AOC} \$=90^{\circ}-\$$ \$ \angle $\mathrm{OCA}=$ \angle $\mathrm{PAC}=$ \angle PHB\$ (the last equality follows from the obvious inscribed nature of the quadrilateral \$АРНС\$). Therefore, ray \$НР\$ cuts off from the right triangle $\$ \mathrm{ABH} \$$ an isosceles triangle $\$ \mathrm{BMH} \$$. Hence, $\$ \mathrm{M} \$$ is the midpoint of $\$ \mathrm{AB} \$$.

![](https://cdn.mathpix.com/cropped/2024_05_06_8c5c78c6462ff090b828g-06.jpg?height=337&width=614&top_left_y=852&top_left_x=-1)