The applet lets you manipulate a quadrilateral, ABCD, so as to maintain the equality of three sides:
BC = CD = AD.
Let E be the point of intersection of the diagonals. An interesting situation occurs when also AE = BE. There are plenty of isosceles trapezoids each of which satisfies both conditions. However, there are quadrilaterals, other than trapezoids that also have these properties:
Clearly, in such cases, CE ≠ DE, for, otherwise, the quadrilateral would be a trapezoid. Interestingly, when this condition holds, invariably ∠AED = 60°.
Following M. Hajja (The Mathematical Gazette, v. 94, n. 531, Nov. 2010), let's introduce angles α, β, ... as shown:
The focus of our interest are triangles ADE and BCE. In these triangles,
AE = BE, AD = BC, ∠AED = ∠BEC.
We are in a position to exploit the SSA situation. There are just two possibilities: Either the two triangles ADE and BCE are congruent, or γ + δ = 180°. The first case is impossible due to the third condition CE ≠ DE. If so, γ + δ = 180° is bound to hold. Summing up the angles in triangles ACD and BCD we get
2α + β + γ = 180°,
α + 2β + δ = 180°.
Adding the two and taking into account that γ + δ = 180° we obtain 3(α + β) = 180°, or
α + β = 60°.
But, by the Exterior Angle Theorem, ε = α + β, and we are done.
Furthermore, again by the Exterior Angle Theorem, ∠ABE + ∠BAE = 60°, implying ∠ABC + ∠BAD = 120°. But this means that the sides BC and AD meet at a 60° angle, which make the quadrilateral ABCD equilic.
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