We have a triangular pyramid (see the picture) in which the side of the base and the height are equal to $a$ and the side walls are isosceles triangles.
The pyramid was cut it with a plane passing through one of the edges of the base such that we obtain two solids of the same volume. Determine the tangent of the angle of inclination of this plane to the base plane.
I made a drawing. Since both solids have the same volume and the base has not changed, the yellow line $h = \frac{H}{2}=\frac{a}{2}$. I do not know what to do next. The blue line in the base is equal to $\frac{a\sqrt{3}}{2}$ because it is a hight of the equilateral triangle of side $a$. I believe that to find the tangent I need to calculate the parts in which the height $h$ of the smaller pyramid divides the height of the base triangle.
I would be grateful for any help.
Give names to the pyramid's vertices. $T$ is the top, the other are $X, Y, Z$ where $Y$ is the rightmost and $Z$ the leftmost. Let $C$ be the centroid of the triangle $X Y Z$, let $M$ and $P$ be the ends of the yellow segment, with $P$ in the base plane, finally let $Q$ be the middle of $X$ and $Z$.
We know that $TC = 2 MP = a$ The intercept theorem tells us that $T Y = 2 M Y$ and that $C Y = 2 P Y =2(CY-CP)$. Hence $2CP = C Y = 2 Q C$ It follows that $QC = C P$, hence $\tan(\alpha) = \frac{a/2}{2 a \sqrt{3}/6} = \frac{\sqrt{3}}{2}$