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The Law of Cosines

[ The ratio of the areas of triangles with a common base or a common height] Class

The diagonals of a convex quadrilateral $ABCD$ are perpendicular and intersect at point $O, AO=2$, $OC=3$. Point $K$ lies on side $BC$, such that $BK: KC=1: 2$. Triangle $AKD$ is equilateral. Find its area.

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Prove that diagonal $B D$ divides segment $A K$ in the ratio 2:1.

## Solution

Let $P$ and $Q$ be the projections of point $K$ onto diagonals $A C$ and $B D$, respectively. Then $K Q=O P=O C / 3=1$.

If $M$ is the point of intersection of segments $A K$ and $Q O$, then from the similarity of triangles $Q M K$ and $O M A$ it follows that $K M$ :
$M A=K Q: A O=1: 2$.

Let $A K=K D=A D=x$. Then $M K=x / 3$. By the cosine theorem, $D M^{2}=M K^{2}+K D^{2}-2 M K \cdot K D \cos 60^{\circ}=$ $7 x^{2} / 9$. Since $A M=2 A K / 3$, then $S_{A M D}=2 / 3 S_{A K D}$, or $\frac{x \sqrt{7}}{3}=\frac{2}{3} \cdot \frac{x^{2} \sqrt{3}}{4}$. From this, $x=\frac{2 \sqrt{7}}{\sqrt{3}}$. Therefore, $S_{A B C}=\frac{x^{2} \sqrt{3}}{4}=\frac{7}{\sqrt{3}}$

![](https://cdn.mathpix.com/cropped/2024_05_06_f7f9d90261bfc416bd71g-22.jpg?height=766&width=600&top_left_y=369&top_left_x=732)

Answer

$\frac{7}{\sqrt{3}}$.

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