Cosine of Sum/Proof 4

Proof

 * [[File:Tri1.PNG]]

$AB$, $AC$, $AE$, and $AD$ are radii of the circle centered at $A$.

Let $\angle BAC = a$ and $\angle DAC = \angle BAE = b$.

By Euclid's First Postulate, we can construct line segments $BD$ and $CE$.

By Euclid's second common notion, $\angle DAB = \angle CAE$.

Thus by Triangle Side-Angle-Side Equality, $\triangle DAB \cong \triangle CAE$.

Therefore, $DB = CE$.

We now assign Cartesian coordinates to the points $B$, $C$, $D$, and $E$:

We use the definition of the distance function on the Euclidean space $\struct {\R^2, d}$ as defined by the Euclidean metric:


 * $\forall x, y \in \R^2: \map d {x, y} = \sqrt {\paren {x_1 - x_2}^2 + \paren {y_1 - y_2}^2}$

where $x = \tuple {x_1, y_1}, y = \tuple {x_2, y_2}$.

Thus:
 * $DB \cong CE \iff \map d {D, B} = \map d {C, E}$

So, plugging in the coordinates of $B, C, D, E$, we get: