Sophism #15: All Triangles are Isosceles

Consider an arbitrary triangle ABC. Let G be the midpoint of AB and let D be the intersection of the midpoint perpendicular of AB with the bisector of ∠C. Drop perpendiculars DE and DF to the sides AC and BC correspondingly.

Now observe that, because CD is a bisector,

Next, consider the triangle ADB. As DG is both a median and a height in this triangle, it must be isosceles, and thus DA = DB. From the equalities DE=DF and DA=DB follows the equality of the right triangles AED and BFD. Consequently:

Finally, add equations (1) and (3) together to obtain:

In other words, ABC is an isosceles triangle.

For other sophisms check out my other posts.