MHB Find the Special Point of a Triangular Pyramid: Proof Explained

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In a triangular pyramid VABC with equal heights from vertex V to the lateral faces, point F in plane ABC is identified as the closest point to V, serving as the altitude. The discussion reveals that F is the intersection of a sphere centered at V, which touches the sides of triangle ABC, indicating that F is the incentre of triangle ABC. The initial suggestion that F might be the median is clarified, as the incentre is the correct identification. The relationship between the pyramid's geometry and triangle properties is emphasized. Thus, F is confirmed as a special point of triangle ABC, specifically the incentre.
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Suppose that the lateral faces VAB, VBC, and V CA of triangular pyramid VABC
all have the same height drawn from V . Let F be the point in plane ABC that is closest
to V , so that VF is the altitude of the pyramid. Show that F is one of the special points
of triangle ABC.

I made the triangular pyramid and I think the special point is the median.
Am I correct?

Thanks.
 
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veronica1999 said:
Suppose that the lateral faces VAB, VBC, and VCA of triangular pyramid VABC
all have the same height drawn from V . Let F be the point in plane ABC that is closest
to V , so that VF is the altitude of the pyramid. Show that F is one of the special points
of triangle ABC.

I made the triangular pyramid and I think the special point is the median.
Am I correct?

Thanks.
Let $d$ be the "height drawn from $V$" of the three lateral faces. The sphere of radius $d$ centred at $V$ touches (tangentially) each of the three sides $BC$, $CA$, $AB$, of the base of the pyramid. Therefore the intersection of the sphere with the plane $ABC$ is the incircle of the triangle $ABC$. The line $VF$ is perpendicular to the plane $ABC$, so that $F$ is the centre of that circle. So I reckon that $F$ is the incentre of the triangle.
 
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