In triangle $ABC, AB>AC$ . $D$ is a point on $AB$ such that $AD=AC$.
Prove that $\,\angle ADC=\frac{\angle B +\angle C}{2}$ and $\angle BCD=\frac{\angle C-\angle B}{2}$.
Solving this problem in Euclidean geometry is very easy. But how can it be solved with the following restrictions?
Parallel postulate (i.e. properties of parallel lines ) cannot be used.
Theorems proved using properties of parallel lines cannot be used.
The problem has to be solved the way euclidean geometry problems are solved. Cartesian Geometry cannot be used.
If this problem can be so solved, then please provide the solution. If not solvable, why cannot be?
I don't think this holds in non-Euclidean geometry. Consider a sphere with $B$ at the South Pole $90^\circ \text{S}$, and $A$ and $C$ very near (and nearly on opposite sides of) the North Pole; say $A = 89^\circ \text{N}, 89^\circ \text{W}$ and $A = 89^\circ \text{N}, 89^\circ \text{E}$. Then $\angle C$ is nearly $180^\circ$ (if you fly due north along the 89th meridian east from $B$ to $C$, you only need to turn slightly to the left to continue from $C$ to $A$, with your closest approach to the North Pole coming on the Prime Meridian). Similarly, $\angle B$ is $178^\circ$, which is also nearly $180^\circ$. But $D$ is somewhere on the 89th meridian east and south of $C$—to be precise, very slightly due north of $87^\circ N, 89^\circ E$. Traveling the path $A \to D \to C$ requires performing a near-complete about-face at $D$, so $\angle ADC$ is very small. This is a counter-example to $\angle ADC = \frac{\angle B + \angle C}{2}$.