Label the edges $BC=a,$ $CA=b,$ $AB=c,$ $DA=d,$ $DB=e,$ $DC=f.$ Construct in the plane $ABC$ triangles $\triangle UAB$ and $\triangle VAC,$ such that $AU=AV=d,$ $UB=e$ and $VC=f.$ Then the projections $M,N$ of $U,V$ on $AB,AC$ coincide with the projections of $D$ on $AB,AC$ and $P \equiv UM \cap VN$ is the orthogonal projection of $D$ on the plane $ABC.$ This follows from the fact that $\triangle UAB$ and $\triangle VAC$ are obtained by rotating the faces $\triangle DAB$ and $\triangle DAC$ about $AB,AC.$ Thus if we construct a right triangle with hypotenuse equal to $UM$ and one leg equal to $MP,$ then the other leg gives the measure of the D-altitude of $ABCD.$