In $\triangle ABC$, $\overline{AA'}, \overline{BB'}$ and $\overline{CC'}$ are concurrent cevians. Points $D,E,F$ are on $\overline{AA'}, \overline{BB'}$ and $\overline{CC'}$ respectively (as in the diagram below) such that $\triangle DEF \approx \triangle ABC$. Prove that the sides of $\triangle DEF$ are parallel to the sides of $\triangle ABC$ respectively or show that this is not the case always.
Note that the points $D,E,F$ are between the vertices of $\triangle ABC$ and the point of contact of the cevians.
This is what I tried :
If $DE \| AB$, then $\frac{PD}{PA} = \frac{PE}{PB} = \frac{DE}{AB}$
If $DF \| AC$, then $\frac{PD}{PA} = \frac{PF}{PC} = \frac{DF}{AC}$
If $EF \| BC$, then $\frac{PE}{PB} = \frac{PF}{PC} = \frac{EF}{BC}$
Equating all $3$ equations yields $\frac{DE}{AB} = \frac{DF}{AC} = \frac{EF}{BC}$, which is true (as the triangles are similar). Thus the line segments are parallel.
Is this proof correct? Are there any counter cases that disproves this proof?
Help is appreciated!
Best Answer
Initial condition: In $\triangle ABC$, cevians AA’, BB’, and CC’ meet at P. D is a point on AP.
Through D, draw DF // AC cutting CC’ at F. Similarly draw DE // AB cutting BB’ at E. Join EF.
The first question is “will $\triangle DEF \sim\triangle ABC$?”
Note that, after the construction, we have $\triangle PAC \sim \triangle PDF$ and this yields $\dfrac {PF}{PC} = \dfrac {PD}{PA}$. Similarly, we have $\dfrac {PE}{PB} = \dfrac {PD}{PA}$.
Then, $\dfrac {PF}{PC} = \dfrac {PE}{PB}$. This means $\triangle PEF \sim \triangle PBC$ because $\angle BPC$ is the common angle. Therefore, $\angle PEF= \angle PBC$ and $\angle PFE=\angle PCB$. By AAA, $\triangle ABC \sim \triangle DEF$. In fact, the $\triangle DEF$ so constructed is the only triangle meeting the given condition.
Finally, EF // BC, a result already obtained in the last paragraph.
Further explanation:
For a particular point D on AP. The lines DE an DF so constructed fixed $\angle EDF$ to match $\angle BAC$.
To simply the picture drawing, we let $\angle BAC = 90^0$. For similarity to occur, we need $\angle EDF = 90^0$ also, so that D is against A.
$\triangle DEF$ is then constructed according to way suggest above.
Suppose there is another triangle $DE’F’ $ constructed with (1) E’ on BP; (2)F’ on PC and $\angle E’DF’ = \angle EDF = \angle BAC = 90^0$.
BY considering $\triangle DFX$ and $\triangle DF’X$, we can see that $\angle DF’X \ne \angle DFX = \angle ACB$. This means $\triangle DE’F’$ so constructed will never be similar to $\triangle ABC$.