Area of a special isosceles spherical triangle

I was asked to calculate the area of the following isosceles triangle.

Let $\triangle ABC$ be an isosceles spherical triangle with $d(A,C)=d(B,C)$. Then $C$ lies on the perpendicular bisector of $AB$. Let $M$ be the midpoint of $AB$. Suppose $d(A,M)=d(-M,C)$. Calculate the area of $\triangle ABC$.

I already made a sketch and observed that $d(A,M)=d(-M,C)=\pi-d(M,C)$.

Let $\alpha$, $\beta$, $\gamma$ denote the interior angles corresponding to $A$, $B$, $C$ respectively.

If I assume without loss of generality that, because $C$ lies on the perpendicular bisector of $AB$, the angle $\gamma$ is divided into two equal angles $\gamma_1$ and $\gamma_2$ belonging to the right triangles $\triangle AMC$ and $\triangle BMC$. Since $$\operatorname{area}(\triangle ABC) = 2 \cdot \operatorname{area}(\triangle AMC) = 2 \cdot \operatorname{area}(\triangle BMC)$$ I only need to calculate $\operatorname{area}(\triangle AMC)$.

Since $\operatorname{area}(\triangle AMC)=\alpha + \gamma_1+ \frac{\pi}{2}-\pi$, I need $\alpha$ and $\gamma_1$. I first tried to calculate $\gamma_1$ using the spherical Law of Cosines, and I got this:

$$\cos\gamma_1=\cos\alpha \cdot \cos\frac{\pi}{2} - \sin\alpha \cdot \sin\frac{\pi}{2} \cdot \cos d(A,M)=-\sin\alpha \cdot \cos\left(\pi-d(M,C)\right) \tag{1}$$

From the spherical Law of Sines, I get: $$\sin\alpha= \frac{\sin d(M,C)}{\sin d(A,C)} \tag{2}$$

But now I do not see how to proceed. I would like to calculate the right-hand sides of $(1)$ and $(2)$, but this does not seem possible with the data I have. Or is this meant to be solved another way? I am not allowed to use Napier's rule since it was not proven in the lecture.

A more geometric approach starts with extending the line segments $MC$ and $MA$ to their meeting point $-M$.
This creates a bi-angle between $M$ and $-M$ with area $\pi.\, AC$ divides this bi-angle into two congruent triangles because $d(A,M) = d(C,-M)$, so $\bigtriangleup$$AMC has area \frac\pi2 and \bigtriangleup$$ABC$ has area $\pi$.