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In the diagram below, $AB$, $EF$ and $DC$ are perpendicular to $BC$. $AEC$ and $BED$ are straight lines. $AB = x$, $EF = h$ and $DC = y$. Then how can $h$ be expressed?

Here is the image:

enter image description here

I drew the shape and used Pythagoras' theorem for the two triangles and made them equal to each other since both equal $h$. Then I used Pythagoras' theorem for the two large triangles and then I was stuck.

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    I believe you have enough rep to put the image in the question, and to know that proper MathJax formatting will help us help you! – 2017-01-25
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    Give your question an appropriate title, please. Make it descriptive and objective. – 2017-01-25
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    Then $h$ is what? – 2017-01-25
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    Is the question incomplete? You say "Then $h$ is" -- what? Is there a list of possibilities you are supposed to choose from? – 2017-01-25

2 Answers 2

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Using similar triangles $$\frac hx=\frac{FC}{BC}\\\frac hy=\frac{BF}{BC}\\ \frac hx + \frac hy = 1\\h=\frac{xy}{x+y}$$

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    why is h/x +h/y =1???? @rossmillikan – 2017-01-26
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    @exchangehelpforuni $BF+FC=BC$ – 2017-01-27
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    @rossmilikan I know why is BC=1?????? – 2017-01-27
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    It is not. But $\frac {BC}{BC}=1$ I am adding the two previous equations. They are already over the common denominator $BC$ – 2017-01-27
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Using only Thales Theorem:

$$\begin{align*} &\text{On }\;\Delta BCD:\;\;\frac{BF}{BC}=\frac hy\implies h\stackrel1=\frac{BF}{BC}y\\{}\\ &\text{On}\;\;\Delta ABC:\;\;\frac{CF}{BC}=\frac hx\implies h\stackrel2=\frac{CF}{BC}x\end{align*}$$

Comparing both expressions for $\;h\;$ above:

$$y\,BF=x\,CF\implies\frac yx=\frac{CF}{BF}$$

and from here

$$h\stackrel1=\frac{BF}{BF+CF}y=\frac1{1+\frac{CF}{BF}}y=\frac1{1+\frac yx}y=\frac{xy}{x+y}$$