Is there a solution to this geometry problem?

Question

The point $G$ is centroid in the triangle $ABC$.

$m(BGC)=90^o$

$|AB|=8 \;cm$

$|CG|=? \, cm$

I know that $|GD|=|BD|=|DC|$ and $|AG|=|BC|=2|GD|$. I'm stuck after this point. I've tried to find the answer by solving system of equations, but it didn't work out. How do I solve this question?

Answer 1

The point $D$ is not uniqely determined. Everything is on a picture:

We have $$4b^2+c^2=16$$ but $$ CG = 2c $$

Answer 2

Perpendicularity of the two medians $m_b$ and $m_c$ is equivalent to $b^2+c^2=5a^2$

Knowing $c=8$ and $b^2+c^2=5a^2$ does not allow us to determine $2a^2+2b^2-c^2$ (that is, the $4m_c^2$, which we would know if we knew $|CG|$)

Consider two triangles, one having $a=\sqrt{20}, b=6, c=8$ and the other $a= \sqrt{45}, b=\sqrt{161}, c=8$.

Both of them satisfy $b^2+c^2=5a^2$, and yet the quantity $2a^2+2b^2-c^2 = 12a^2 - 3c^2$ is different for them, since their $a$'s differ.

Answer 3

I'll start by recapping, with explanation, some things OP already knows. (I believe OP made a typo in asserting $|AG|=|BC|=\frac12|GD|$. The "$\frac12$" should (obviously?) be a "$2$".)

Since $\overline{AD}$ is a median, the condition that $\angle BGC$ is right tells us that $D$ is the midpoint of the hypotenuse of right triangle $\triangle BGC$; therefore, $D$ is the circumcenter of that triangle: $|BD|=|GD|$. Moreover, we know that a centroid trisects each median: $|AD|=3\,|GD|$.

Thus, the perpendicular medians tell us this about $D$: $$|AD| = 3\,|BD|$$

Now, there's a whole family of points $D$ whose distance from $A$ is three-times their distance from $B$: they live on the circle with diameter $\overline{D_1D_2}$ where $D_1$ and $D_2$ are the two easily-located points on $\overleftrightarrow{AB}$ (one between $A$ and $B$, the other on the "far side" of $B$) satisfying that property.

Every point $D$ on this circle (except perhaps $D_1$ and $D_2$ themselves, depending upon the reader's tolerance for degenerate cases) gives rise to a viable $\triangle ABC$.

However, the target distance, $|CG|$, varies, from a minimum of $0$ (when $D=D_1$) to a maximum of $|AB|$ (when $D=D_2$). Consequently, there is no unique solution to the problem described. $\square$