The solution equates the ratio of area of the triangles with the ratio of the sides. How is this even possible? I thought the general rule is that the ratio of the area of the triangle is equal to the SQUARE of the ratio of the sides.
2026-05-06 11:48:50.1778068130
Geometry problem with diagram, with the concept of ratio of areas.
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The two triangles $\triangle AFE$ and $\triangle BFE$ have the same height (using $F$ as the top corner and $AE$ and $BE$ as the bases), so the ratio between their areas equals the ratio between their bases. The other three equalities follow a similar reasoning.
In general, for any two triangles, the ratio between their areas is the product of the ratio of their bases with the ratio of their heights. What you're thinking of, when you're thinking of squares, is probably for similar triangles. In that case, the ratio of the bases is equal to the ratio of the heights. In your problem, however, the ratio of the heights is equal to $1$.