$AD$,$BE$,$CF$ are concurrent at $I$ , which is the incenter of $\triangle DEF$. $DE$ cut $CF$ at $G$ and then $(F,G/I,C)=-1$ $~$ $\Longrightarrow$ $~$ $\angle IEC=\pi$/2 Likewise, $~$ $\angle IDB=\angle IFA=\pi$/2 and then it's done.

gold46 wrote: Let $\angle AFE = \angle BFD = x$ $\angle BDF = \angle EDC=y$ $\angle DEC = \angle FEA = z$ $\angle ABE = a$ $\angle BEC = b$ Obviously $x+y+z=180$.(*) $\angle ABC=a+b=180-x-y$(1) $\angle AEB = 180 - \angle BEC \implies a-b=x-y$ (2) (1) , (2) $\implies a=90-y \implies \angle AEB=180 - 180 +x+y -90 +z=^{(*)}90$ Similarly $\angle ADC = \angle CFB=90$

See Here for this same problem discussed an yr earlier.

Let <BDF=<CDE=x, <CED=<AEF=y, <AFE=<BFD=z.Applying the law of sines to BDF,CDE and AFE we get BD/BF=sinz/sinx, CD/CE=siny/sinx, AE/AF=sinz/siny. So BD/CD*CE/AE*AF/BF=1.....(1). We draw a line l parallel to BC passing through A. DF and DE (extended) meet l at points X and Y. Note that BDF is similar to AXF and so BD/BF=AX/AF. Similarly CD/CE=AY/AE. Substituting these back in (1) we get AX=AY. In triangles AXD and AYD we have AX=AY, <AXD=<AYD=x and DX=DY. Therefore triangles AXD and AYD are conguent and so <XAD=<YAD=90 degrees.Hence XY parallel to BC implies <BDA=<CDA=90 degrees. BE and CF can be proved altitudes in a similar way. BYE

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We see that $\angle AFE=\angle BFD=\gamma, \angle BDF=\angle CDE=\alpha, \angle CED=\angle AEF=\beta$ since $\alpha+\beta+\gamma=180^\circ$ $ABDE, AFDC$ and $FBCE$ are cyclic since $\angle EDB+\angle EAB=\angle CAF+\angle CDF=\angle ECB+\angle EFB=180^\circ$ $\angle ADF=\angle ECF=\angle EBA=\angle EDA$. Now: $\angle CDA=\angle CDE+\angle EDA=\angle BDF+\angle ADF=\angle ADB$ Since $\angle CDA+\angle ADB=180^\circ \implies \angle CDA=\angle ADB=90^\circ$ From cyclic quadrilateral $ABDE: \angle ADB=\angle AEB=\angle BEC=90^\circ$ From cyclic quadrilateral $AFDC: \angle CDA=\angle CFA=\angle CFB=90^\circ$. That is all