Let the two transversals to $ABC$ be $DEF$ and $D_1E_1F_1,$ where we assume that $D$ and $E_1$ lie outside the triangle, and $D,D_1\in BC; E,E_1\in CA; F,F_1\in AB.$ Then we need to show that $FD\cdot F_1E_1=FE\cdot F_1D_1.$ Let the two lines passing through $H$ and parallel to $AB, AC$ meet $BC$ at points $M,N.$ Then note that the pencil $(HF,HF_1, HB, HM)$, on a counter-clockwise rotation of $+\dfrac{\pi}{2},$ gives the pencil $(HE_1, HE, HN, HC).$ Therefore, we see that $\dfrac{BF}{CE}=\dfrac{BF_1}{CE_1}\ \ (*).$ [asy][asy] import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -6.97, xmax = 7.08, ymin = -1.54, ymax = 7.78; /* image dimensions */ pen qqzzqq = rgb(0,0.6,0); pen aqaqaq = rgb(0.63,0.63,0.63); pen ffttcc = rgb(1,0.2,0.8); pen qqqqff = rgb(0,0,1); pen ffttww = rgb(1,0.2,0.4); pen ffttcc = rgb(1,0.2,0.8); pen aqaqaq = rgb(0.63,0.63,0.63); pen ffwwqq = rgb(1,0.4,0); pen wwzzff = rgb(0.4,0.6,1); pen qqzzqq = rgb(0,0.6,0); pen ffttcc = rgb(1,0.2,0.8); pen ffffff = rgb(1,1,1); /* draw figures */ draw((0,4.53)--(-2,0), qqzzqq); draw((-2,0)--(6.34,0), aqaqaq); draw((6.34,0)--(0,4.53), ffttcc); draw((1.22,3.66)--(-4,0), qqqqff); draw((-2.42,6.25)--(1.96,0), ffttww); draw((0,4.53)--(-2.42,6.25), ffttcc); draw((-2,0)--(-6.47,0), aqaqaq); draw((0,4.53)--(3.17,0), ffwwqq); draw((0,4.53)--(-6.47,0), wwzzff); draw((-1.24,0)--(0,2.8), qqzzqq); draw((0,2.8)--(3.92,0), ffttcc); /* dots and labels */ dot((0,4.53),linewidth(2pt) + dotstyle); label("$A$", (0.08,4.61), NE * labelscalefactor); dot((-2,0),linewidth(2pt) + dotstyle); label("$B$", (-2,-0.2), S * labelscalefactor); dot((6.34,0),linewidth(2pt) + dotstyle); label("$C$", (6.43,-0.2), S * labelscalefactor); dot((0,2.8),linewidth(2pt) + dotstyle); label("$H$", (0.08,2.88), N * labelscalefactor); dot((-4,0),linewidth(2pt) + dotstyle); label("$D$", (-3.98,-0.2), S * labelscalefactor); dot((1.22,3.66),linewidth(2pt) + dotstyle); label("$E$", (1.3,3.74), NE * labelscalefactor); dot((-1.1,2.03),linewidth(2pt) + dotstyle); label("$F$", (-1.16,2.15), NW * labelscalefactor); dot((-2.42,6.25),linewidth(2pt) + dotstyle); label("$E_1$", (-2.4,6.32), NE * labelscalefactor); dot((-0.47,3.47),linewidth(2pt) + dotstyle); label("$F_1$", (-0.65,3.38), NW * labelscalefactor); dot((1.96,0),linewidth(2pt) + dotstyle); label("$D_1$", (2.05,-0.2), S * labelscalefactor); dot((-1.24,0),linewidth(2pt) + dotstyle); label("$M$", (-1.17,-0.2), S * labelscalefactor); dot((3.92,0),linewidth(2pt) + dotstyle); label("$N$", (4.01,-0.2), S * labelscalefactor); dot((-6.47,0),linewidth(2pt) + dotstyle); label("$X$", (-6.43,-0.2), S * labelscalefactor); dot((3.17,0),linewidth(2pt) + dotstyle); label("$Y$", (3.26,-0.2), S * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); shipout(bbox(ffffff,Fill)); [/asy][/asy] Now, observe that, if lines through $A$ and parallel to $DEF$ and $D_1E_1F_1$ meet $BC$ at $X$ and $Y,$ then \[\frac{BF}{AB}=\frac{BD}{BX},\ \frac{BF_1}{AB}=\frac{BD_1}{BY}, \ \frac{AC}{EC}=\frac{XC}{DC}, \ \frac{AC}{E_1C}=\frac{YC}{D_1C}.\] Therefore, \[\begin{aligned}\left|\frac{(B,C,D,X)}{(B,C,D_1,Y)}\right|&=\dfrac{\frac{DB}{DC}\cdot \frac{XC}{XB}}{\frac{D_1B}{D_1C}\cdot\frac{YC}{YB}}\\&=\frac{DB\cdot XC\cdot D_1C\cdot YB}{DC\cdot XB\cdot D_1B\cdot YC}\\&=\frac{DB}{XB}\cdot \frac{YB}{D_1B}\cdot \frac{XC}{DC}\cdot\frac{D_1C}{YC}\\&=\frac{BF}{AB}\cdot \frac{CE}{CA}\cdot \frac{F_1B}{AB}\cdot\frac{E_1C}{AC}\\&=\frac{BF\cdot CE}{F_1B\cdot E_1C}\\&=1 \ \ [\text{from} (*)].\end{aligned}\] So, we see that \[(AB,AC,AD,AX)\cap \overleftrightarrow{DEF}=(AB,AC,AD,AY)\cap \overleftrightarrow{D_1E_1F_1};\] Leading to \[\frac{FD}{FE}=\frac{F_1D_1}{F_1E_1}\implies FD\cdot F_1E_1=FE\cdot F_1D_1.\] $\Box$ P.S This problem is actually a well-known theorem (I can't seem to recall the name).

Suppose, in the triangle $ ABC $, $ A_1B_1C_1,A_2B_2C_2 $ are the two perpendicular lines passing through the orthocenter $ H $ of $ ABC$. $ A_1,A_2 $ lie on $ BC $, $ B_1,B_2 $ lie on $ AC $ and $ C_1,C_2 $ lie on $ AB $. We will prove that there exists a spiral similarity that takes $ A_1,B_1,C_1 $ to $ A_2,B_2,C_2 $, in other words, if $ A_3,B_3,C_3 $ are the circumcenters of $ \odot HA_1A_2,\odot HB_1B_2, \odot HC_1C_2 $, then $ A_3,B_3,C_3 $ are collinear. Suppose, $ X$ is the foot of the perpendicular from $ H $ to $ B_3C_3 $. Now note that, $ \angle C_3HB_3=180-\angle BAC$. So $ \angle B'XC'=180-2\angle BAC $ where $ AA',BB',CC' $ are the altitudes of $ ABC$. So $ X $ lies on the nine-point circle of $ ABC $. Similarly define $ Y,Z $ for $ A_3B_3,A_3C_3 $. But note that, $ X,Y,Z $ lie on the circles with diameter $ HA_3,HB_3,HC_3 $ too. So $ X\equiv Y\equiv Z $. So $ A_3,B_3,C_3 $ are collinear. So done.

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