geometrytriangles
The exterior angle theorem for triangles states that the sum of “The measure of an exterior angle of a triangle is equal to the sum of the measures of the two interior angles that are not adjacent to it; this is the exterior angle theorem. The sum of the measures of the three exterior angles (one for each vertex) of any triangle is $360$ degrees.” How can we prove this theorem?
So, for the triangle above, we need to prove why $\angle CBD = \angle BAC + \angle BCA$.
in $ \triangle ABC$, $\angle B=90^\circ$, $AD=DC$
A vectorial approach would be quite lean and effective.
Given two faces of the pyramid, sharing the common edge $V P_n$, and containing the contiguous base points $P_{n-1}$ and $P_{n+1}$, the dihedral angle between these two faces would be the angle made by the two vectors ($t_m, t_p$), normal to the common edge and lying on the respective face.
Clearly, that will be also the angle made by the normal vectors to the faces, provided that one is taken in the inward, and the other in the outward direction.
That is, by the right-hand rule, $$ {\bf n}_{\,m} = \mathop {P_{\,n} P_{\,n - 1} }\limits^ \to \; \times \;\mathop {P_{\,n} V}\limits^ \to \quad \quad {\bf n}_{\,p} = \mathop {P_{\,n} P_{\,n + 1} }\limits^ \to \; \times \;\mathop {P_{\,n} V}\limits^ \to $$
Then the dihedral angle $\alpha$ will be simply computed from the dot product $$ \cos \alpha = {{{\bf n}_{\,m} \; \cdot \;{\bf n}_{\,p} } \over {\left| {{\bf n}_{\,m} } \right|\;\;\left| {{\bf n}_{\,p} } \right|}} $$
Best Answer
In the triangle $\triangle ABC$, we know that $ABD$ is a straight line. So $\angle ABC = 180^{\circ} - \angle CBD$. From the angle sum property of triangles we can infer that $\angle BAC + \angle ABC + \angle BCA = 180^{\circ}$ or $\angle ABC = 180^{\circ}-(\angle BAC + \angle BCA)$. Therefore:
$$\angle ABC = 180^{\circ} - \angle CBD = 180^{\circ} - (\angle BAC + \angle BCA)$$ $$\Rightarrow - \angle CBD = -(\angle BAC + \angle BCA)$$ $$\Rightarrow - \angle CBD \times -1 = -(\angle BAC + \angle BCA) \times -1$$ $$\Rightarrow \angle CBD = \angle BAC + \angle BCA$$