In Fig. 13., since _aR_ is by construction parallel to _AB_ in
Fig. 12., and _TV_ is by construction in Problem III. also parallel to
_AB_—
∴ _aR_ is parallel to _TV_,
∴ _abR_ and _TbV_ are alternate triangles,
∴ _aR_ ∶ _TV_ ∷ _ab_ ∶ _bV_.
Again, by the construction of Fig. 13., _aR′_ is parallel to _MV_—
∴ _abR′_ and _MbV_ are alternate triangles,
∴ _aR′_ ∶ _MV_ ∷ _ab_ ∶ _bV_.
And it has just been shown that also
_aR_ ∶ _TV_ ∷ _ab_ ∶ _bV_—
∴ _aR′_ ∶ _MV_ ∷ _aR_ ∶ _TV_.
But by construction, _aR′_ = _aR_—
∴ _MV_ = _TV_.