If in a tetrahedron ABCD the heights are congruent and A is projected on the (BCD) plane in the orthocenter, ABCD is a regular tetrahedron

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Tianna Deleon · Accepted Answer

Step 1Let ABCD is tetrahedron with congruent heights. Volume of tetrahedron is equal       1    3        h    i        S    i   , where       h    i   and       S    i   are corresponding heights and faces areas. As       h    i   are equal, so       S    i   are also equal.Let       A    1   is projection of A on the (BCD) plane, and       A    1   is the orthocenter of triangle BCD.Consider altitude BE of triangle BCD.  A      A    1    ⊥  (  B  C  D  )  ,      A    1    E  ⊥  C  D  ⇒  A  E  ⊥  C  DThen AE is altitude of triangle ACD. Area of faces BCD and ACD are equal   C  D  ⋅  B  E      /    2 and   C  D  ⋅  A  E      /    2 and are equal to each other. Then   A  E  =  B  E . Then triangles CEB and CEA are congruent and triangles DEB and DEA are congruent. Then   A  C  =  B  C and   A  D  =  B  D .Using the same method with altitude CF passing through       A    1   , one can obtain   A  B  =  B  C ,   A  D  =  C  D . Using the same method with altitude DG passing through       A    1   , one can obtain   A  C  =  C  D ,   A  B  =  B  D . Then   A  B  =  A  C  =  B  C  =  C  D  =  A  D  =  B  D . Then tetrahedron ABCD is regular.

If in a tetrahedron ABCD the heights are congruent and

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