<Solid-2025-255>, hint

Let E be a point on CD and HE is perpendicular to CD.

Since HH' is orthogonal to plane ACD, A, B, H, H' E are collinear

Suppose AE>BE and  rotate ACD so that ACD and BCD lie on the same plane

Let M be the midpoint of CD. Since MG:GB=1:2=MG':G'A, G,G', F are collinear

F is the orthogonal projection of G, G' onto CD

GF=1/3 BE(h), G'F=1/3AE(h')  GF/G'F=h/h'

HE=CE cot CHE=CE cot CDB=CE (ED/BE)=(CE ED)/h

H'E=CE cot CH'E=CE cot CDA=CE(ED/AE)=(CE ED)h'

GF/G'F=h/h'=H'E/HE