Let $A(0,0),B(1,0),C(1,1),D(0,1)$. Choose $E(a,0), F(1,1-a)$. Equation of $EC : y = \frac{1}{1-a}(x-a)$, line $l \bot EC$, equation of $l : y=(a-1)(x-1)$. Solving: $N(\frac{a^{2}-a+1}{a^{2}-2a+2},\frac{(a-1)^{2}}{a^{2}-2a+2})$. Point $G(0,1-a)$, equation of $FG : y=1-a$. Point $P(a^{2}-a+1,1-a)$. Slope of $DP : m_{DP}= - \frac{a}{a^{2}-a+1}$. Equation of $NF : y -(1-a) = (a^{2}-a+1)(x-1)$. Point $T(\frac{a}{a^{2}-a+1}+1,1)$. Slope of $BT : m_{BT}= \frac{a^{2}-a+1}{a}$. Conclusion: $DP \bot BT$.

Note that $ AENG $ is cyclic $ \implies BN.BG=BE.BA=BF.BC $, so $ \angle GFC = 90^{\circ} $ i.e, $ GF \parallel AB $ and points $ N,G,D,C,F $ are concyclic. Suppose that $ DP $ meets $ BC $ at $ X $. We will show that $ CT=CX $. By the similar triangles $ \triangle XFP, \triangle XCD $, $ CX=\frac{CD.FX}{FP} $. By the similar triangles $ \triangle XFP, \triangle DGP $, $ FX=\frac{PF.GD}{PG} $. Combine to get $ CX = \frac{CD.GD}{PG}=\frac{CB.CF}{PG} $. By the similar triangles $ \triangle NPF, \triangle NCT $, $ CT=\frac{NC.PF}{NP} $. Therefore, we need to show $ \frac{CB.CF}{PG}=\frac{NC.PF}{NP} $ $ \iff $ $ CB.CF = \frac{CN.PF.PG}{PN} $. But, we have $ PF.PG=PN.PC $, so we only need to show $ CB.CF=CN.CP $. This is true since $ BNPF $ is cyclic ( $ \angle BNP=\angle BFP =90^{\circ} $ ). Now, $ \angle BDT = 45^{\circ}= \angle XTC $ $ \implies TX \perp BD $, and $ BX \perp TD $, so $ X $ is the orthocenter of $ \triangle BDT $ $ \implies DX \perp BT $, i.e $ DP \perp BT $.

Dear Mathlinkers, 1. I used http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=127915 for proving that GF // AB 2. in order to prove that NEF and NBD are similar I consider the similatities of NGP and NCB, NGD and NCT, then we are done... Sincerely Jean-Louis

Let $X = BC \cap DP$. Since $\angle ABG= \angle BCE$ we get $\triangle ECB \cong \triangle GBA$. Thus $AG=BE=BF$. Moreover $ABFG$ is a rectangle. Furthermore $G, N, F, C$ and $D$ are cyclic. $\triangle GPD \sim \triangle FPX$. $\Longrightarrow \frac{XF}{GD} = \frac{FP}{GP}$ $\Longrightarrow \frac{XF+GD}{GD} = \frac{FP+GP}{GP}$ $\Longrightarrow XC= \frac{GD \cdot GF}{GP}$. On the other hand $\triangle NGP \sim \triangle NCB$ $\Longrightarrow \frac{CN}{GN} = \frac{CB}{GP}$ Finally $\triangle CFT \sim \triangle NGC$ $\Longrightarrow \frac{CT}{CN} = \frac{CF}{GN}$ $\Longrightarrow CT = \frac{CF \cdot CN}{GN} = \frac{GD \cdot GF}{GP} = XC$ Thus $\triangle DXC \cong \triangle BTC$. Furthermore $\angle XDC = \angle TBC$. We conclude $DX \perp BT$

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