Hello, mathbuoy!
Did you make a sketch?
Two towers with heights $\displaystyle a$ and $\displaystyle b$ meters are 100 meters apart.
What is the height of the intersection point of the lines
joining the top of each tower to the base of the other? Code:
P *
|*
| *
| *
| *
| *
a | * * R
| *T * |
| * |
| * | | b
| * h| * |
| * | *|
Q * - - - * - * S
100-x U x
The towers are: .$\displaystyle PQ = a\,\text{ and }\,RS = b$
The lines cross at $\displaystyle T$, .$\displaystyle h = TU$
Let $\displaystyle US = x\text{, then }QU = 100-x$
Since $\displaystyle \Delta TUS \sim \Delta PQS\!:\;\;\frac{h}{x} = \frac{a}{100} \quad\Rightarrow\quad h \:=\:\frac{ax}{100}$ .[1]
Since $\displaystyle \Delta TUQ \sim \Delta RSQ\!:\;\;\frac{h}{100-x} = \frac{b}{100}\quad\Rightarrow\quad h \:=\:\frac{b(100-x)}{100}$ .[2]
Equate [1] and [2]: .$\displaystyle \frac{ax}{100} \:=\:\frac{b(100-x)}{100} \quad\Rightarrow\quad x \:=\:\frac{100b}{a+b} $
Substitute into [1]: .$\displaystyle h \:=\:\frac{a}{100}\left(\frac{100b}{a+b}\right) \quad\Rightarrow\quad \boxed{h\;=\; \frac{ab}{a+b}} $
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An observation . . .
Divide top and bottom by $\displaystyle ab\!:\;\;h \;=\;\frac{\dfrac{ab}{ab}}{\dfrac{a}{ab} + \dfrac{b}{ab}} \;=\;\frac{1}{\dfrac{1}{a} + \dfrac{1}{b}}$
The answer is one-half the harmonic mean of $\displaystyle a\text{ and }b.$