∴ _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_. III. ANALYSIS OF PROBLEM XV. We proceed to take up the general condition of the second problem, before left unexamined, namely, that in which the vertical distances _BC′_ and _AC_ (Fig. 6. page 13), as well as the direct distances _TD_ and _TD′_ are unequal. In Fig. 6., here repeated (Fig. 76.), produce _C′B_ downwards, and make _C′E_ equal to _CA_. [Illustration: Fig. 76.] Join _AE_. Then, by the second Corollary of Problem II., _AE_ is a horizontal line. Draw _TV_ parallel to _AE_, cutting the sight-line in _V_. ∴ _V_ is the vanishing-point of _AE_. Complete the constructions of Problem II. and its second Corollary. Then by Problem II. _ab_ is the line _AB_ drawn in perspective; and by its Corollary _ae_ is the line _AE_ drawn in perspective. From _V_ erect perpendicular _VP_, and produce _ab_ to cut it in _P_. Join _TP_, and from _e_ draw _ef_ parallel to _AE_, and cutting _AT_ in _f_. Now in triangles _EBT_ and _AET_, as _eb_ is parallel to _EB_ and _ef_ to _AE_;—_eb_ ∶ _ef_ ∷ _EB_ ∶ _AE_. But _TV_ is also parallel to _AE_ and _PV_ to _eb_. Therefore also in the triangles _aPV_ and _aVT_, _eb_ ∶ _ef_ ∷ _PV_ ∶ _VT_. Therefore _PV_ ∶ _VT_ ∷ _EB_ ∶ _AE_. And, by construction, angle _TPV_ = ∠ _AEB_. Therefore the triangles _TVP_, _AEB_, are similar; and _TP_ is parallel to _AB_. Now the construction in this problem is entirely general for any inclined line _AB_, and a horizontal line _AE_ in the same vertical plane with it. So that if we find the vanishing-point of _AE_ in _V_, and from _V_ erect a vertical _VP_, and from _T_ draw _TP_ parallel to _AB_, cutting _VP_ in _P_, _P_ will be the vanishing-point of _AB_, and (by the same proof as that given at page 17) of all lines parallel to it. [Illustration: Fig. 77.] Next, to find the dividing-point of the inclined line. I remove some unnecessary lines from the last figure and repeat it here, Fig. 77., adding the measuring-line _aM_, that the student may observe its position with respect to the other lines before I remove any more of them. Now if the line _AB_ in this diagram represented the length of the line _AB_ in reality (as _AB_ _does_ in Figs. 10. and 11.), we should only have to proceed to modify Corollary III. of Problem II. to this new construction. We shall see presently that _AB_ does not represent the actual length of the inclined line _AB_ in nature, nevertheless we shall first proceed as if it did, and modify our result afterwards. In Fig. 77. draw _ad_ parallel to _AB_, cutting _BT_ in _d_.