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If a trapezoid has bases of length $a,b$, find the length of the segment that is parallel to the bases and divides the trapezoid into $2$ equal areas.

To make it clear, $FH\parallel CD\parallel AB$ and $A_{ABHF}=A_{FHDC}$. The bases are known, i.e. $AB=a$ and $CD=b$. We are asked to find $FH$ in terms of $a$ and $b$.

Basically this seems like an easy problem, but for some reason I can't solve it. As you can see on the diagram below, I have added a perpendicular line $AE\perp CD$ and a parallel line $AI\parallel BD$. I know that, e.g., $\triangle AFK\sim \triangle ACI$ and hence $\frac{AJ}{AE}=\left(\frac{A_{AFK}}{A_{ACI}}\right)^2$, but that doesn't seem to help me out.

I am not sure if the fact that I've added these additional perpendiculars and parallel lines will help me out. That's just an idea I thought of when solving this. So I need some help now. Thanks.

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  • $\begingroup$ possible duplicate of Trapezoid Root Mean Square $\endgroup$ Commented Feb 6, 2014 at 23:22

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Hint: Let AE be $h$. The area is then $\frac 12h(a+b)$ If JE$=d$, we want to evaluate FH. Note the similar triangles AFK and ACI and that KH$=a$. You want $\frac 12(FH+b)d=\frac 14h(a+b)$ There is enough info here to solve.

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