[ Area of a triangle (using two sides and the angle between them).] Four points lying on the same circle

In trapezoid $A B C D$, diagonals $A C$ and $B D$ are perpendicular to each other, $\angle B A C=\angle C D B$. The extensions of the lateral sides $A B$ and $D C$ intersect at point $K$, forming angle $A K D$, which is $30^{\circ}$. Find the area of triangle $A K D$, if the area of the trapezoid is $P$.
Prove that trapezoid $ABCD$ is isosceles.

## Solution

Let $AD > BC$. Then points $K$ and $A$ lie on opposite sides of line $BC$. Since $\angle BAC = \angle CDB$, a circle can be circumscribed around trapezoid $ABCD$. Therefore, the trapezoid is isosceles. Consequently, triangles $AKD$ and $BKC$ are also isosceles.

Since

$$
\angle BAC = \angle BAD - \angle CAD = 75^{\circ} - 45^{\circ} = 30^{\circ},
$$

triangle $ACK$ is isosceles, with $CK = AC$. Then

$$
S_{\triangle BKC} = \frac{1}{2} CK^2 \sin 30^{\circ} = \frac{1}{4} CK^2
$$

and since

$$
P = S_{ABCD} = \frac{1}{2} AC \cdot BD = \frac{1}{2} AC^2 = \frac{1}{2} CK^2
$$

we have $S_{\Delta BKC} = \frac{1}{2} P$. Therefore,

$$
S_{\triangle AKD} = \frac{1}{2} P + P = \frac{3}{2} P
$$

If $AD < BC$, then points $K$ and $A$ lie on the same side of line $BC$. In this case, reasoning similarly, we get that

$$
S_{\triangle AKD} = \frac{1}{2} P
$$

## Answer

$\frac{3}{2} P$ or $\frac{1}{2} P$