Prove that all three bisecting planes of the dihedral angles of a trihedral angle intersect along one straight line.

Question

Prove that all three bisecting planes of the dihedral angles of a trihedral angle intersect along one straight line. I attempted like this we can take one triangle at first as all its bisectors will intersect at one common point which can be proved using ceva's theorem all the following triangles are similar so in every triangle exists a point which is perpendicular to three faces i want to show tha all these point lie in a single line

REFER TO THIS A trihedral angle is a geometric term used to describe the space formed by the intersection of three half-planes. It consists of three planar faces that meet at a common point called the vertex. Each pair of adjacent faces forms a dihedral angle.

A dihedral angle, on the other hand, is the angle between two intersecting planes. It is formed by the intersection of two planar faces of a polyhedron or the intersection of two half-planes in a trihedral angle.

In the context of a trihedral angle, the bisecting planes of the dihedral angles refer to the planes that divide each dihedral angle into two equal angles. These bisecting planes intersect at a common line known as the axis of the trihedral angle. The statement you mentioned asserts that in a trihedral angle, all three bisecting planes intersect along a single straight lineIn the given image show that line EG exists and is unique

Answer 1

If $u_1, u_2, u_3$ are the unit normal vectors pointing towards the inside of the trihedral angle, then, the bisector of the dihedral angle between planes (1) and (2) has a normal equal to

$ n_{12} = u_1 - u_2 $

Similarly

$n_{23} = u_2 - u_3 $

and

$ n_{13} = u_1 - u_3 $

Note that $n_{13} = n_{12} + n_{23} $

Now the direction vector of the line of intersection between the first bisecting plane and the second plane is given by the cross product of $n_{12} $ and $n_{23}$, i.e

$ D_1 = n_{12} \times n_{23} $

Similarly,

$ D_2 = n_{23} \times n_{13} = n_{23} \times \bigg( n_{12} + n_{23} \bigg) = n_{23} \times n_{12} $

And

$ D_3 = n_{13} \times n_{12} = \bigg( n_{12} + n_{23} \bigg) \times n_{12} = n_{23} \times n_{12} $

Hence,

$ D_1 = - D_2 = - D_3 $

That is, the line of action of all direction vectors $D_1, D_2, D_3$ is the same. And since the bisecting planes pass through the trihedral vertex. Then the intersection of the bisecting planes is the same line.