Solution of triangles in Non-Euclidean geometry with restrictionsIn triangle A B C , A B &gt; A C. D is a point on AB such that A D = A C.Prove that ∠ A D C = ∠ B + ∠ C 2 and ∠ B C D = ∠ C − ∠ B 2 .Solving this problem in Euclidean geometry is very easy. But how can it be solved with the following restrictions?1) Parallel postulate (i.e. properties of parallel lines ) cannot be used.2) Theorems proved using properties of parallel lines cannot be used.3) The problem has to be solved the way euclidean geometry problems are solved. Cartesian Geometry cannot be used.If not solvable, why cannot be?

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Solution of triangles in Non-Euclidean geometry with restrictionsIn triangle   A  B  C  ,  A  B  &amp;gt;  A  C. D is a point on AB such that   A  D  =  A  C.Prove that     ∠  A  D  C  =            ∠      B      +      ∠      C        2   and   ∠  B  C  D  =            ∠      C      −      ∠      B        2  .Solving this problem in Euclidean geometry is very easy. But how can it be solved with the following restrictions?1) Parallel postulate (i.e. properties of parallel lines ) cannot be used.2) Theorems proved using properties of parallel lines cannot be used.3) The problem has to be solved the way euclidean geometry problems are solved. Cartesian Geometry cannot be used.If not solvable, why cannot be?

Makai Lang · Accepted Answer

Step 1I don&#039;t think this holds in non-Euclidean geometry. Consider a sphere with B at the South Pole       90    ∘    S, and A and C very near (and nearly on opposite sides of) the North Pole; say   A  =      89    ∘    N  ,      89    ∘    W and   A  =      89    ∘    N  ,      89    ∘    E. Then   ∠  C is nearly       180    ∘   (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).Step 2Similarly,   ∠  B is       178    ∘  , which is also nearly       180    ∘  . But D is somewhere on the 89th meridian east and south of C—to be precise, very slightly due north of       87    ∘    N  ,      89    ∘    E. Traveling the path   A  →  D  →  C requires performing a near-complete about-face at D, so   ∠  A  D  C is very small. This is a counter-example to   ∠  A  D  C  =            ∠      B      +      ∠      C        2  .

In triangle ABC, AB>AC . D is a point on AB such that AD=AC. Prove that angle ADC= (angle B+ angle C)/2 and angle BCD=(angle C - angle B)/2.

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