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 $\left({\R^2, d}\right)$ as defined by the Euclidean metric:


 * $\forall x, y \in \R^2: d \left({x, y}\right) = \sqrt {\left({x_1 - x_2}\right)^2 + \left({y_1 - y_2}\right)^2}$

where $x = \left({x_1, y_1}\right), y = \left({x_2, y_2}\right)$.

Thus:
 * $DB \cong CE \iff d \left({D, B}\right) = d \left({C, E}\right)$

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