Given segments $a$ and $b$. Using a compass and a straightedge, construct the segment $\sqrt{a b}$.

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Use the property of the height of a right triangle drawn from the vertex of the right angle.

## Solution

Construct segments $A B$ and $B C$ ($B$ between $A$ and $C$) on an arbitrary straight line. On segment $A C$ as a diameter, construct a circle. Through point $B$, draw a line perpendicular to $A C$. Let $D$ be one of the points of intersection of the drawn line with the circle. Since $\angle A D C=90^{\circ}$, segment $D B$ is the height of the right triangle $A D C$, drawn from the vertex of the right angle. Therefore, $D B^{2}=A B \cdot B C=a b$, from which $D B=$ $\sqrt{a b}$.

Construct segments $A B$ and $B C$ ($B$ between $A$ and $C$) on an arbitrary straight line. On segment $A C$ as a diameter, construct a circle. Through point $B$, draw a line perpendicular to $A C$. Let $D$ be one of the points of intersection of the drawn line with the circle. Since $\angle A D C=90^{\circ}$, segment $D B$ is the height of the right triangle $A D C$, drawn from the vertex of the right angle. Therefore, $D B^{2}=A B \cdot B C=a b$, from which $D B=$ $\sqrt{a b}$.

Construct segments $A B$ and $B C$ ($B$ between $A$ and $C$) on an arbitrary straight line. On segment $A C$ as a diameter, construct a circle. Through point $B$, draw a line perpendicular to $A C$. Let $D$ be one of the points of intersection of the drawn line with the circle. Since $\angle A D C=90^{\circ}$, segment $D B$ is the height of the right triangle $A D C$, drawn from the vertex of the right angle. Therefore, $D B^{2}=A B \cdot B C=a b$, from which $D B=$ $\sqrt{a b}$.

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