Prove the concurrency of lines $BI$, $CH$ and $FG$.

Question

For the triangle $ABC$, we have $DE||BC$, $FG||BC$, and $HI||BC$. Also, point $H$ is the inverse of point $D$ wrt line $FG$ in the rectangle $DEIH$.

Prove that $BI$, $CH$, and $FG$ are concurrent at point $J$.

My attempts:

I think there is pretty easy proof using Thales's theorem around point $J$. But I'm more interested in using the concurrency of the line segments constructed from similar points of two parallel triangles. So, I'm trying to find a point on the line $FG$...

Answer 1

In the figure, $PAQ$ and $LJM$ are lines perpendicular to $HI$ and $BC$.

Note that

$$\frac{AP}{AQ}=\frac{AR}{AQ}=\frac{DE}{BC}=\frac{HI}{BC} \tag{1}$$

Also $\triangle JHI \sim \triangle JCB \implies $ $$\frac{LJ}{JM}=\frac{HI}{BC} \tag{2}$$

$(1)$ and $(2)$ $\implies$

$$\frac{LJ}{JM}= \frac{PA}{AQ} \tag{3}$$

Since $PQ=LM$, $(3) \implies $

$$MJ=AQ$$

i.e. $J$ has the same height from $BC$ as that of $A$.

Therefore $J$ lies on $FG$.

Answer 2

We reproduce in $\mathrm{Fig.\space 1}$ the sketch given by OP after removing the lines $BI$ and $CH$. We also added some new lines, segments, and points. First, we located the points of meeting $M$ and $N$ between $FG$ with $DH$ and $EI$ respectively. After drawing the segment-pairs $\{BN,\space CM\}$ and $\{MI,\space NH\}$, the points of intersection of each pair were denoted as $S$ and $R$ respectively. Finally, lines $BH$ and $CI$ were drawn and extended to intersect with the line drawn through $S$ and $R$ at $Q$ and $P$ respectively.

Since $HMNI$ is a rectangle, we have, $$MR=RI=NR=NH\qquad\rightarrow\qquad \dfrac{MR}{RI}=\dfrac{NR}{RH}=1.\tag{1}$$

Since $MN$ is parallel to $BC$, $\triangle SNM$ and $\triangle SBC$ are similar. Therefore, we shall write, $$\dfrac{MS}{SC}=\dfrac{NS}{SB}=k, \quad\text{where} \space k\space\text{is some real number}.\tag{2}$$

Line $PRS$ cuts the three sides of $\triangle MCI$, while line $QRS$ does the same to the sides of $\triangle NHB$. We apply Menelaus’ theorem to both these instances to obtain, $$\dfrac{MR}{RI}\times\dfrac{CS}{SM}\times\dfrac{IP}{PC}=1\qquad\text{and}\qquad \dfrac{BS}{SN}\times\dfrac{NR}{RH}\times\dfrac{HQ}{QB}=1.\tag{3}$$

From (1), (2), and (3), we can deduce that, $$\dfrac{IP}{PC}=\dfrac{HQ}{QB}=k.\tag{4}$$

Since $HI$ is parallel to $BC$, $P$ and $Q$ must be the one and same point in order for the equality (4) to hold. With that we can conclude that the line through $S$ and $R$ goes through the point of intersection of extended $BH$ and $CI$.

We have redrawn $\mathrm{Fig.\space 1}$ to depict the fact we just proved above. By the way, we named point of intersection of $BH$ and $CI$ as $P$. We also included the segments $BI$ and $CH$ in $\mathrm{Fig.\space 2}$. Please pay your attention to $\triangle MCI$ and $\triangle NHB$.

The corresponding pairs of sides of these two triangles, i.e., $\{BH,\space CI\}$, $\{HN,\space IM\}$, and $\{NB,\space MC\}$, intersect at $P$,$R$, $S$, which are collinear. This means that $\triangle MCI$ and $\triangle NHB$ are perspective from a line. According to Desargues’ theorem, if two triangles are perspective from a line, they are perspective from a point as well. Therefore, the lines joining the corresponding pairs of vertices of these two triangle, i.e., $BI$, $HC$, and $NM$, are concurrent at $J$, which is called the perspective center.