Let $D$ be a point inside an isosceles triangle $\triangle{ABC}$ where $AB=AC$. If $\angle{ADB} > \angle{ADC}$, prove $DC > DB$.
This problem can be solved using the rotation method. This method is discussed in the book %%HREF%%Geometry Technique%%http://www.mathallstar.org/Home/Home/35-books/94-book-geometry-techniques%%. Let's rotate $\triangle{ADB}$ counter-clockwise by $60^\circ$ to $\triangle{AD'C}$. Because $AB=AC$, point $B$ will move to point $C$ after such a rotation. $$\triangle{ABD} \cong \triangle {ACD'} \implies AD=AD'\quad\text{and}\quad \angle{AD'C} = \angle{ADB} > \angle{ADC}$$ Connect $D$ and $D'$. Because $AD=AD'$, we have $\angle{ADD'} = \angle{AD'D}$. Hence, $\angle{DD'C} > \angle{D'DC} \implies DC > D'C = DB$
