Why angle form by a secant and tangent outside a circle is half the intercepted arc?

I know that according to Inscribed Angle Theorem that the measure of an angle will be half the intercepted arc's measure.

I know that the measure of the angle formed from tangent and secant is equal to one-half the difference of the measures of intersected arcs i.e Angle = $\frac 12(\text{long - short})$

But I don't understand why a secant and tangent outside a circle angle is also half of the intercepted arc?

According this Tangent-Secant Proof I saw, it says $\angle CAE = \dfrac12 AC.$ Why?

Image source: http://jwilson.coe.uga.edu/emt668/EMAT6680.2003.fall/Nichols/6690/Webpage/Day%209.htm

The sketch given below might help you to find out why $m\measuredangle CAE = \dfrac{1}{2}m\overset{\huge\frown}{AC}.$

$\underline{\text{Added at the Request of OP}}$

As you can see from the sketch, we have added a point $O$ to mark the center of the circle. We join $O$ to existing points $A$ and $C$ to obtain two radii to facilitate the following explanation. For brevity, we let $\measuredangle CDA =\alpha$

Using inscribed angle theorem, we shall state, $$\measuredangle COA = 2\measuredangle CDA = 2\alpha. \tag{1}$$

$\triangle COA$ is an isosceles triangle, because $OA$ and $OC$ are two radii. Therefore, we have $$\measuredangle OAC = \measuredangle ACO. \tag{2}$$

Since the sum of the three internal angles of a triangle is equal to two right angles or $180^0$, we can write, $$\measuredangle OAC + \measuredangle ACO + \measuredangle COA = 180^0. \tag{3}$$

We eliminate $\angle ACO $ and $\angle COA $ from (3) by substituting the values given in (1) and (2) and get, $$2\measuredangle OAC + 2\alpha = 180^0 \qquad\longrightarrow\qquad \measuredangle OAC = 90^0 - \alpha \tag{4}$$

Since any given radius of a circle always makes an angle of $90^0$ with the tangent drawn to the circle at the point of intersection between the radius and the circumference, $\measuredangle OAE = 90^0$. Now, using (4), we are in a position to express $\angle CAE$ as, $$\measuredangle CAE =\measuredangle OAE - \measuredangle OAC = 90^0 - \left(90^0 - \alpha \right) = \alpha.$$

According to this and (1), we have, $$\measuredangle CAE = \measuredangle CDA = \frac{1}{2}\measuredangle COA = \dfrac{1}{2}m\overset{\huge\frown}{AC}$$