Folkor

In a convex quadrilateral $A B C D: \angle B A C=20^{\circ}, \angle B C A=35^{\circ}, \angle B D C=40^{\circ}, \angle B D A=70^{\circ}$. Find the angle between the diagonals of the quadrilateral.
We will prove that point $D$ is the center of the circumcircle of triangle $ABC$. This can be done in various ways.

The first method. Describe a circle around triangle $ABC$ and extend segment $BD$ to intersect this circle at point $K$ (Fig. a). Since $\angle BKC = \angle BAC = 20^\circ$, then $\angle KCD = \angle BDC - \angle DKC = 20^\circ$ (angle $BDC$ is the external angle for triangle $KDC$). Therefore, $DC = DK$.

Similarly, since $\angle BKA = \angle BCA = 35^\circ$ and $\angle BDA = 70^\circ$, then $\angle KAD = 35^\circ$, so $DK = DA$.

The second method. On ray $AD$, mark point $M$ such that segment $DM = DB$ (Fig. b). Then $\angle DBM = \angle BMD = \frac{1}{2} \angle BDA = 35^\circ = \angle BCA$, so points $A, B, C$, and $M$ lie on the same circle.

Similarly, marking point $P$ on ray $CD$ such that $DP = DB$, we find that point $P$ lies on the same circle. Point $D$ is equidistant from points $B, M$, and $P$, so it is the center of the obtained circle.
![](https://cdn.mathpix.com/cropped/2024_05_06_1b164f828743b46f847cg-12.jpg?height=430&width=1162&top_left_y=412&top_left_x=456)

The third method. The center of the circumcircle of obtuse triangle $ABC$ lies in the same half-plane relative to line $AC$ as point $D$ (Fig. a, b). It is the intersection of two loci: one from which segment $BC$ is seen at an angle $\alpha = 2 \angle BAC = 40^\circ$ and one from which segment $AB$ is seen at an angle $\beta = 2 \angle BCA = 70^\circ$.

In the specified half-plane, these loci are arcs of circles that have a unique common point. By the condition, from point $D$, these same segments are seen at the same angles, so point $D$ coincides with the center of the circumcircle of triangle $ABC$.

Let $T$ be the intersection point of the diagonals of quadrilateral $ABCD. \angle DBA = \frac{1}{2}(180^\circ - \angle BDA) = 55^\circ$, angle $BTC$ is the external angle for triangle $BTA$, so $\angle BTC = \angle TAB + \angle ABT = 75^\circ$.

## Answer

$75^\circ$.