In Fig. 8. omit the lines _CD_, _C′D′_, and _DS_; and, as here in
Fig. 75., from _a_ draw _ad_ parallel to _AB_, cutting _BT_ in _d_;
and from _d_ draw _de_ parallel to _BC′_.
[Illustration: Fig. 75.]
Now as _ad_ is parallel to _AB_—
_AC_ ∶ _ac_ ∷ _BC′_ ∶ _de_;
but _AC_ is equal to _BC′_—
∴ _ac_ = _de_.
Now because the triangles _acV_, _bc′V_, are similar—
_ac_ ∶ _bc′_ ∷ _aV_ ∶ _bV_;
and because the triangles _deT_, _bc′T_ are similar—
_de_ ∶ _bc′_ ∷ _dT_ ∶ _bT_.
But _ac_ is equal to _de_—
∴ _aV_ ∶ _bV_ ∷ _dT_ ∶ _bT_;
∴ the two triangles _abd_, _bTV_, are similar, and their angles
are alternate;
∴ _TV_ is parallel to _ad_.
But _ad_ is parallel to _AB_—
∴ _TV_ is parallel to _AB_.