# Proof: Isosceles Triangles in a Quadrilateral

In my last post, I noted that it’s possible to create an isosceles trapezoid from four isosceles triangles, but I wasn’t sure if there was a way to construct a quadrilateral from isosceles triangles such that the quadrilateral was neither a rectangle nor an isosceles trapezoid. Now I know that it is not.

Let’s reconsider the basic quadrilateral with its diagonal segments marked:

Using 1, 2, 3, and 4 to represent distinct lengths of the spoke segments e, f, g, and h, we have a total of fifteen possible configurations: 1111, 1112, 1121, 1122, 1123, 1211, 1212, 1213, 1221, 1222, 1223, 1231, 1232, 1233, and 1234. However, many of these are duplicates. In reality, we have seven distinct configurations for e, f, g, and h:

1. All congruent (1111)
2. All but one congruent (1112)
3. Two pairs of consecutive congruent segments (1122)
4. Two pairs of alternate congruent segments (1212)
5. One pair of consecutive congruent segments (1123)
6. One pair of opposite congruent segments (1213)
7. No congruent segments (1234)

Now consider $$\Delta ABC$$, with subtriangles $$\Delta ABE$$ and $$\Delta BEC$$. If $$g \not\cong h$$, then either $$g \cong a$$ or $$h \cong a$$. If $$g \cong a$$, then $$\angle BEA \cong \angle EBA$$ and therefore $$m\angle BEA < 90^{\circ}$$. Likewise, if $$h \cong a$$, then $$\angle BEA \cong \angle EAB$$ and $$m\angle BEA < 90^{\circ}$$. By similar reasoning, if $$e \not\cong h$$, then $$m\angle BEC < 90^{\circ}$$.

However, $$m\angle BEC + m\angle BEA = 180^{\circ}$$. So either $$g \cong h$$ or $$e \cong h$$. This is true for all four spokes. In other words, every spoke must be consecutive with a congruent segment. This rules out cases 2 and 4 to 7 above, where at least one segment does not have a congruent neighbor. Therefore the only two configurations that allow for four isosceles triangles are 1 (all spokes are congruent) and 3 (two pairs of consecutive congruent segments).

When all spokes are congruent, we have a square.

With two pairs of consecutive congruent segments, we have two logical possibilities. Let us say that $$e \cong f$$ and $$g \cong h$$. Then either $$d \cong b$$ or $$d \not\cong b$$. But comparing $$\Delta DEA$$ to $$\Delta CEB$$, we know that $$f \cong e$$, $$h \cong g$$, and $$\angle DEA \cong \angle CEB$$, so $$\Delta DEA \cong \Delta CEB$$ and hence $$d \cong b$$.

Furthermore, comparing $$\Delta DEC$$ to $$\Delta AEB$$, since $$e : f :: g : h$$ (i.e, $$1 = 1$$) and $$\angle DEC \cong \angle AEB$$, $$\Delta DEC$$ is similar to $$\Delta AEB$$. Because the triangles are isosceles, $$\angle EDC \cong \angle ECD$$ and $$\angle EAB \cong \angle EBA$$; because the triangles are similar, then, $$\angle CDE \cong \angle EBA$$. Those angles are alternate interior on transversal $$\overleftrightarrow{BD}$$, so $$c || a$$.

In this case, then, we have a quadrilateral where one pair of sides is parallel and the other pair is congruent. This is either a parallelogram (specifically, having congruent diagonals, a rectangle) or an isosceles trapezoid.

Thus, if the four non-overlapping triangles created by the diagonals of a quadrilateral are all isosceles, the quadrilateral must be either a rectangle or an isosceles trapezoid.

Clio Corvid

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