Prove ∠PAB=∠CAQ in Tetrahedron ABCD

[anemone]

The insphere and the exsphere opposite to the vertex $D$ of a (not necessarily regular) tetrahedron $ABCD$ touch the face $ABC$ in the points $P$ and $Q$, respectively. Show that $\angle PAB=\angle CAQ$.

 

[jvicens]

Thank you for your interesting question about the relationship between the insphere and exsphere in a tetrahedron. I am happy to provide you with an explanation of this relationship.

First, let us define the terms insphere and exsphere. The insphere of a tetrahedron is the largest sphere that is tangent to all four faces of the tetrahedron. The exsphere, on the other hand, is the largest sphere that is tangent to three faces of the tetrahedron and to the plane containing the fourth face.

Now, let us consider the tetrahedron $ABCD$ as described in the forum post. Since the insphere and exsphere are tangent to the face $ABC$, they must touch this face at a point on each of their surfaces. Let us call these points $P$ and $Q$, respectively.

Now, since the insphere is tangent to all four faces of the tetrahedron, it is also tangent to the faces $ABD$, $ACD$, and $BCD$. This means that the point $P$ is equidistant from the three edges $AB$, $AC$, and $BC$. Similarly, since the exsphere is tangent to three faces and the plane containing the fourth face, the point $Q$ is equidistant from the edges $AB$, $AC$, and $BC$.

From this, we can see that the points $P$ and $Q$ are equidistant from the edges $AB$, $AC$, and $BC$, and therefore they must lie on the angle bisector of the angle $BAC$. This means that $\angle PAB=\angle CAQ$, as the forum post suggests.

I hope this explanation helps to clarify the relationship between the insphere and exsphere in a tetrahedron. If you have any further questions, please do not hesitate to ask.