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A line segment parallel to the base of an equilateral triangle is drawn passing through the two legs such that the area of the smaller triangle and the trapezoid formed are equal. What is the ratio of the height of the smaller triangle to the height of the trapezoid.

I tried assigning values for the sides, but just couldn't find the value of the two heights.

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  • $\begingroup$ What goes wrong? I wouldn't expect the answer to be anything terribly pleasant, but the computation should be straightforward. Note: since everything scales, you might as well suppose the height of the original triangle to be $1$. Might simplify the work. $\endgroup$ – lulu Feb 20 at 11:15
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Hint:

  1. Use Basic Proportionality Theorem.
  2. Areas of two similar triangles are in the ratio of the length of their corresponding sides squared.
  3. Construct Altitude $CL$ which intersects $XY$ at $M$ and $AB$ at $L$.
  4. Corresponding Altitudes of two similar triangles are in the ratio of the corresponding sides.

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$$\dfrac{\text{ar}(\Delta CXY)}{\text{ar}(AXYB)}=1 \implies \dfrac{\text{ar}(\Delta CXY)}{\text{ar}(\Delta CBA)}=\dfrac{1}{2}=\dfrac{CM^2}{CL^2}$$ Now you simply need to find $CM/ML$ from here. Can you proceed?

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