$BE,BC,CD$ form side lengths of a triangle

Question

In triangle $ABC$, point $D$ lies on $AB$ and $E$ on $AC$ such that $AD=AE$. Prove that $BE,BC,CD$ form side lengths of a triangle.

By triangle inequality, this is equivalent to proving that $BE+BC>CD$, $BE+CD>BC$, and $BC+CD>BE$. By symmetry, we only need to show the first two inequalities. Let $BE$ and $CD$ intersect at $X$. Then $BE+CD>BX+CX>BC$.

How can we prove that $BE+BC>CD$?

Answer 1

Observe that $BC+BD>CD$ and $BD+AD<BE+AE$ or $BD<BE$. Thus, $BC+BE>BC+BD>CD$