如何解答这道题？---**Question Number:** 10
**Question Stem:**
在三角形 $ABC$ 中，点 $D$ 为 $BC$ 边之中点，点 $E$ 为 $CA$ 边之中点。线段 $AD$ 与 $BE$ 互相垂直，下图中标记出其交点 $G$，这个点称作三角形 $ABC$ 的重心，且具有 $AG = 2DG$、$BG = 2EG$ 的性质。请问 $\frac{BC^2 + AC^2}{AB^2}$ 之值是什么？
**Mathematical Expression to evaluate:**
$\frac{BC^2 + AC^2}{AB^2}$
**Diagram Description:**
* **Type:** Geometric figure (Triangle).
* **Main Elements:**
* **Vertices:** Points labeled A, B, C forming a triangle.
* **Lines:** Segments AB, BC, CA forming the sides of the triangle. Segments AD and BE, which are medians.
* **Points:** Point D is the midpoint of BC. Point E is the midpoint of CA. Point G is the intersection of AD and BE.
* **Annotations:** Point G is labeled at the intersection. Points D and E are labeled on BC and CA respectively. The vertices A, B, C are labeled.
* **Relationships:** AD is a line segment connecting vertex A to the midpoint D of BC. BE is a line segment connecting vertex B to the midpoint E of CA. AD and BE intersect at G.
* **Implicit Information from text:** AD is perpendicular to BE. G is the centroid. AG = 2DG, BG = 2EG.
**Options:**
No options are provided in the image.