10,11
[Application of trigonometric formulas (geometry) $]$
[ Identical transformations (trigonometry). ]

![](https://cdn.mathpix.com/cropped/2024_05_06_1560f9d9b3af2f34b251g-59.jpg?height=43&width=223&top_left_y=2307&top_left_x=0)
In a convex quadrilateral $A B C D: A C \perp B D, \angle B C A=10^{\circ}, \angle B D A=20^{\circ}, \angle B A C=40^{\circ}$. Find $\angle B D C$.
Let $K$ and $M$ be the points of intersection of the line $CB$ with the line $AD$ and the circumcircle of triangle $ACD$, respectively (see figure). Then

$\angle MDA = \angle MCA = 10^{\circ}$, so $DM$ is the bisector of angle $KDB$. Also note that $\angle ABD = 50^{\circ}$, $\angle CBD = 80^{\circ}$, hence $\angle KBA = 50^{\circ}$, which means $BA$ is the bisector of angle $KBD$.

Thus, $I$ - the intersection point of the bisectors $BA$ and $DM$ - is the incenter of triangle $KBD$. $\angle BID = 120^{\circ}$, so $\angle BKD = 60^{\circ}$. Therefore, quadrilateral $KAIM$ is cyclic, and $KI$ is the bisector of angle $AKM$. Consequently, $\angle ACD = \angle AMD = \angle AMI = \angle AKI = 30^{\circ}$, from which $\angle BDC = 60^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_1560f9d9b3af2f34b251g-60.jpg?height=752&width=712&top_left_y=1&top_left_x=667)

## Answer

$60^{\circ}$.