Finding the third vertex of a isosceles triangle given the vertices of the base and knowing that the medians are perpendicular

Question

Full problem: The points A(0, c) and B(c, 0) are the vetrices of the base of an isosceles triangle. In addition, the medians AD and BE are perpendicular to each other. What are the coordinates of the third vertex?

Even if I figure out the lengths of all sides, medians included, how do I find the coordinates of the vertex?

Answer 1

Hint: the third vertex $C$ will lie on the perpendicular bisector of $AB$, so it will be of the form $(d,d)$. The midpoint $F$ of $AB$ is $(\frac{c}{2},\frac{c}{2})$. The centroid $G$ of the triangle is one third of the way from $F$ to $C$. Now figure out what $d$ must be for $AGB$ to be right.

Answer 2

1. Extend $\overline{AD}$ to $\overline{AE}$ such that $AD = DE$.
2. Bisect $\overline{DE}$ at point $G$.
3. Draw line $\overleftrightarrow{BG}$
4. Intersect line $\overleftrightarrow{BG}$ and $\overleftrightarrow{CD}$ at point F.