Let $ABCS$ be a regular three-sided pyramid, with base $ABC,$ where the side edges are twice as long as the edge of the base, and let $E$ be the point of the side edge $BS$. The angle of inclination of the plane $ACE$ to the base is equal to half of the angle of inclination of the side to the base. Determine the ratio in which the plane $ACE$ divides the volume of the pyramid without trigonometry.
The volume of this pyramid is $V=\dfrac{a^2\sqrt{3}H}{12}$ and we find $H$ from:
$H^2=(2a)^2-(\frac{a\sqrt{3}}{3})^2,$ so $H=\dfrac{11\sqrt{3}a}{3}.$ Therefore, $V=\dfrac{11}{12}a^3.$
We know $\theta=2\varphi.$
We can find the height $h$ of the side edge, but I'm not sure do I need it.
If we denote the point of intersection on $AC$ with $D,$ $ED$ would be the height of $ACE.$ Also, $ED$ would be bisector of the angle $\angle BDS.$ I don't know what else to do.
Let us work in triangle $(T)=SBB'$ where $B'$ is the midpoint of $AC$.
Let $E'$ be the orthogonal projection of $E$ onto the horizontal plane. Let us show how to compute $EE'$ ; knowing it, one has immediately the volume of the little pyramid.
As $BE$ is the angle bisector of angle $B$ in $(T)$, an ill-known (and useful) theorem says that the foot of an angle bisector determines on the opposite side (here the side $SB$) two line segments whose ratio if equal to the ratio of the corresponding lengths of opposite sides, i.e., here, with your notations :
$$\frac{ES}{EB}=\frac{B'S}{B'B}=\frac{h}{a\sqrt{3}/2}\tag{1}$$
giving
$$\frac{EB}{SB}=\frac{a\sqrt{3}/2}{a\sqrt{3}/2+h}\tag{2}$$
Besides, if we call $S'$ the projection of $S$ onto the basis (which is the centroid of $ABC$), we have the proportionnality relationship :
$$\frac{EB}{SB}=\frac{EE'}{SS'} \iff EE'=SS' \times \frac{EB}{SB}\tag{3}$$
But, in the expression above, $SS'$ is known and $\frac{EB}{SB}$ as well (see (2)).
Done ! (you need only compute $h$ as a function of $a$).