Carnot theorem helps here. The condition triangle $ABC$ is perpendicular to triangle $DEF$ is equivalent to: $AE^2-AF^2+BF^2-BD^2+CD^2-CE^2=0$ (1) And the desired property is equivalent to: $EA^2-EC^2+DC^2-DB^2+FB^2-FA^2=0$ But it is obvious the same as (1), so we get the desired one. (Q.E.D)

That directly follows from Carnot's Theorem , but it's an extension , maybe the same solution but still posting $(BD^2-DC^2)+(EC^2-EA^2)+(AF^2-FB^2)=0$ and then multiply both sides by $(-1)$ to get the concurrency condition for the other triple of perpendiculars . $\blacksquare$

From the trigonometric ceva be can easily get the conclusion