Can you solve this tricky viral triangle puzzle?

This is a classic angle-chasing trick. The key is to notice two hidden right angles. Since: $$\angle A=50^\circ,\quad \angle B=60^\circ,\quad \angle C=70^\circ,$$ and $E$ is chosen so that: $$\angle EBC=20^\circ,$$ we get: $$\angle ABE=40^\circ.$$ In triangle $ABE$: $$\angle AEB=180^\circ-50^\circ-40^\circ=90^\circ.$$ So $BE$ is perpendicular to $AC$. Similarly, since $D$ is chosen so that: $$\angle DCB=30^\circ,$$ we have: $$\angle ACD=40^\circ.$$ In triangle $ACD$: $$\angle ADC=180^\circ-50^\circ-40^\circ=90^\circ.$$ So $CD$ is perpendicular to $AB$. Now look at quadrilateral $BDEC$. It has: $$\angle BEC=90^\circ,\qquad \angle BDC=90^\circ.$$ That means $B,D,E,C$ lie on a circle with diameter $BC$. Angles standing on the same chord $DB$ are equal, so: $$\angle DEB=\angle DCB=30^\circ.$$ Therefore: $$\boxed{30^\circ}.$$ The construction feels mysterious until you spot that $BE$ and $CD$ are just two altitudes.