Example 45. An isosceles trapezoid is circumscribed around a circle. Find the ratio of the area of the trapezoid to the area of the circle, given that the distance between the points of tangency of the circle with the lateral sides of the trapezoid is to the radius of the circle as $\sqrt{3}: 1$.

Given (Fig. 49): $A B C D$-trapezoid, $|A B|=|C D|, \omega(0,|O K|)-$ circle inscribed in the trapezoid, $K$ and $M$ - points of tangency with segments $A B$ and $C D,|K M|:|O K|=\sqrt{3}: 1$.

Find: $S_{A B C D}: S_{\omega}$.
S o l u t i o n. As in Example 44, this is a problem of type $M$. We will solve it similarly by introducing an auxiliary parameter. Let $|O K|=R$. We will calculate the area of trapezoid $A B C D$, making some additional constructions beforehand. From point $B$, drop $[B L] \perp [A D]$, and from point $K-[K N] \perp [A D]$. Additionally, draw the radius $O E$ to the point of tangency $E$ of the circle $\omega$ with side $B C$. Let $[K M] \cap [O E]=P$.

Since $[B L] \parallel [K N]$, then $\triangle A K N \sim \triangle A B L$ and, consequently, $\widehat{A K N}=\widehat{A B L}$. Since $\widehat{A K N}=\widehat{A B L}$, and $\widehat{N K M}=\widehat{L B C}=90^{\circ}$, then $\widehat{A K M}=\widehat{A B C}$, hence $[B C] \parallel [K M]$.

Since $[O E] \perp [B C]$, then $[O E] \perp [K M]$ and, consequently, $\triangle O K P$ is a right triangle.

Since $\angle A B L \cong \angle P K O$ (angles with mutually perpendicular sides), then $\triangle A B L \sim \triangle O K P$ and

$$
S_{A B C D}=\frac{|B C|+|A D|}{2} \cdot|B L|
$$

Since in an isosceles trapezoid inscribed around a circle, $|B C|+|A D|=2|A B|$ and $|B L|=2 R$, then $S_{A B C D}=|A B| \cdot 2 R$.

From the similarity of $\triangle A B L$ and $\triangle P K O$ it follows that

$$
\frac{|B L|}{|A B|}=\frac{|K P|}{|K O|}=\frac{\sqrt{3}}{2}
$$

from which

$$
|A B|=\frac{2|B L|}{\sqrt{3}}=\frac{4 R \sqrt{3}}{3}
$$

We obtain:

$$
S_{A B C D}=\frac{8 R^{2} \sqrt{3}}{3}
$$

Further, since $S_{\omega}=\pi R^{2}$, then $\frac{S_{A B C D}}{S_{\omega}}=\frac{8 R^{2} \sqrt{3}}{3 \pi R^{2}}=\frac{8 \sqrt{3}}{3 \pi}$.

I n v e s t i g a t i o n. By the meaning of the problem, $\frac{S_{A B C D}}{S_{\omega}}>1$. We will show that the found ratio of areas satisfies this inequality. Indeed, since $8 \sqrt{3}>8 \cdot 1.7=13.6$, and $3 \pi > 9.42$, then $\frac{8 \sqrt{3}}{3 \pi} > 1$.

Thus, the required ratio of areas is $\frac{8 \sqrt{3}}{3 \pi}$.

## Problems for Independent Solving

67. In triangle $A B C$, $|A B|=26 \mathrm{~cm}, |B C|=30 \mathrm{~cm}, |A C|=$ $=28 \mathrm{~cm}$. From vertex $B$, the height $B H$ and the bisector $B D$ are drawn. Find the area of triangle $B H D$.
68. The legs of a right triangle are 2 and 3 cm. Find the bisector drawn from the vertex of the right angle.
69. Inside an equilateral triangle with side $a$, a point $M$ is taken. Find the sum of the distances from this point to the sides of the triangle.
70. Prove that the sum of the distances from any point taken inside a regular polygon to the lines passing through the sides is a constant value.

72

71. Prove that in any triangle $\frac{1}{r}=\frac{1}{h_{a}}+$ $+\frac{1}{h_{b}}+\frac{1}{h_{c}}$, where $r$ is the radius of the inscribed circle in the triangle, and $h_{a}, h_{b}, h_{c}$ are the heights of the triangle.
72. Prove that the area of a triangle is equal to $\frac{2}{3} m_{a} m_{b} \times$ $\times \sin \left(\widehat{m_{a} m_{b}}\right)$, where $m_{a}$ and $m_{b}$ are the medians of the triangle, and $\left(\widehat{m_{a}} m_{b}\right)$ is the angle between these medians.
73. One of the medians of a triangle, the area of which is $16 \mathrm{~cm}^{2}$, is 6 cm, and the other is 4 cm. Prove that these medians are mutually perpendicular.
74. In an isosceles trapezoid, the midline is equal to $a$, and the diagonals are mutually perpendicular. Find the area of the trapezoid.
75. The area of an isosceles trapezoid inscribed around a circle is $S$, and the height of the trapezoid is half its lateral side. Find the radius of the inscribed circle.
76. The area of an isosceles trapezoid inscribed around a circle is $Q$. Find the lateral side of the trapezoid, the acute angle of which is $30^{\circ}$.
77. On the sides of triangle $A B C$, squares $A B D E, B C P T$, and $A C M N$ are constructed, not overlapping each other. The area of triangle $A B C$ is $9 \mathrm{~cm}^{2}$. Find the areas of triangles $D B T, C P M$, and $A E N$.
78. In a right triangle, one of the acute angles is $\alpha$, and the distance from the vertex of the other acute angle to the center of the inscribed circle is $l$. Find the area of the triangle.

79(2). Through a point taken inside a triangle, three lines are drawn parallel to the sides of the triangle. These lines divide the area of the triangle into six parts, three of which are triangles with areas $S_{1}, S_{2}$, and $S_{3}$. Find the area of the given triangle.

80. The areas of the triangles formed by the segments of the diagonals of a trapezoid with its bases are $S_{1}$ and $S_{2}$. Find the area of the trapezoid.

81(1). In triangle $A B C$, the angle at vertex $C$ is $60^{\circ}$, and the radius of the circle circumscribed around this triangle is $2 \sqrt{3}$. On side $A B$, a point $D$ is taken such that $|A D|=2|D B|$ and $|C D|=2 \sqrt{2}$. Find the area of triangle $A B C$.

82. Chord $A B$ subtends an arc of a circle, the length of which is $\frac{1}{3}$ of the length of the entire circle. On this arc, a point $C$ is taken, and on chord $A B$, a point $D$ is taken. Given $|A D|=2 \mathrm{~cm}, |B D|=1 \mathrm{~cm}$, and $|C D|=\sqrt{2}$. Find the area of triangle $A B C$.

83(10). Find the area of a triangle inscribed in a circle, if the ends of one of its sides, which is 20 cm long, are at distances of 25 and 16 cm from the tangent drawn through the opposite vertex.

84(7). The larger base of a trapezoid is the diameter of the circle circumscribed