5. In an isosceles trapezoid \(ABCD\) with bases \(AD\) and \(BC\), perpendiculars \(BH\) and \(DK\) are drawn from vertices \(B\) and \(D\) to the diagonal \(AC\). It is known that the feet of the perpendiculars lie on the segment \(AC\) and \(AC=20\), \(AK=19\), \(AH=3\). Find the area of trapezoid \(ABCD\).

(10 points)
Solution. Note that right triangles $D K A$ and $B H C$ are similar, since

$\angle B C H=\angle D A K$. Let $D K=x, B H=y$. Due to similarity $\frac{D K}{K A}=\frac{B H}{H C}, \frac{x}{19}=\frac{y}{17}$. On the other hand, $C D$

$=\mathrm{AB}$ and by the Pythagorean theorem

$$
C D^{2}=D K^{2}+K C^{2}=x^{2}+1, A B^{2}=B H^{2}+H A^{2}=y^{2}+9
$$

From this, $x^{2}-y^{2}=8$. Since $y=\frac{17 x}{19}$, substituting $y$ into the last equation, we get $\mathrm{x}=19 / 3, \mathrm{y}$ $=17 / 3, S_{A B C D}=A C \cdot(D K+B H) / 2=20 \cdot(19 / 3+17 / 3)=120$.

Answer: 120.