Example 29. In a trapezoid, the bases of which are respectively equal to $a$ and $b(a>b)$, a line parallel to the bases is drawn through the point of intersection of the diagonals. We will find the length of the segment of this line, cut off by the lateral sides of the trapezoid.

Given (Fig. 33): $A B C D$ - trapezoid, $|A D|=a,|B C|=b$, $[A C]$ and $[B D]$ - diagonals, $(P Q) \|(A D), M \in(P Q), M=[A C] \cap$ $\cap[B D], P \in[A B], Q \in[C D]$.

To find: $|P Q|$.
S o l u t i o n. We will find $|P Q|$ by direct calculation. From $\triangle B C M \sim \triangle A M D$ we have:

$$
\frac{|B M|}{|M D|}=\frac{b}{a}
$$

from which, by the property of the derived proportion, we can write:

$$
\frac{|B M|}{|B M|+|M D|}=\frac{b}{a+b}, \text { or } \frac{|B M|}{|B D|}=\frac{b}{a+b}
$$

From $\triangle B P M \sim \triangle B A D$ we have:

$$
\frac{|P M|}{|A D|}=\frac{|B M|}{|B D|}=\frac{b}{a+b}
$$

Thus,

$$
\frac{|P M|}{a}=\frac{b}{b+a}
$$

Similarly, from $\triangle D Q M \sim \triangle D C B$ we get:

$$
\frac{|M Q|}{|B C|}=\frac{a}{a+b}
$$

T. e.

$$
\frac{|M Q|}{b}=\frac{a}{a+b}, \text { or } \frac{|M Q|}{a}=\frac{b}{a+b}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_318d1c34bf6979fc9790g-053.jpg?height=617&width=671&top_left_y=231&top_left_x=344)

Fig. 34
By adding the equalities (29.1) and (29.2) term by term, we find that

$$
|P Q|=\frac{2 a b}{a+b}
$$

I n v e s t i g a t i o n. By the meaning of the problem, if $b < a$, then $2 a > a + b$, i.e., $2 a b > (a+b) b$, from which

$$
\frac{2 a b}{a+b}>b, \text { or }|P Q|>b
$$

Similarly, we get that $|P Q| > a$.

If $a = b$, then $|P Q| = a = b$.

If $a \neq b$, then $|P Q| < a + b$.

Thus, for any positive values of $a$ and $b$,

$$
|P Q| = \frac{2 a b}{a+b}
$$

and, in particular, if $a = b$, then $|P Q| = a = b$.

If $x > p, \\ x > q, \\ x \geqslant p+q,$

T. e. $x \geqslant p+q$.

Then $x^{2} \geqslant p^{2}+2 p q+q^{2}$, or $\frac{4}{3}\left(p^{2}+q^{2}+p q\right) \geqslant p^{2}+2 p q+$ $+q^{2}$, from which

$$
(p-q)^{2} \geqslant 0
$$

52

Thus, for any positive values of $p$ and $q$,

$$
|A B|=\frac{2 \sqrt{3\left(p^{2}+q^{2}+p q\right)}}{3}
$$

and, in particular, if $p=q$, then $|A B|=2 p$.