The question is:-
A $\Delta ABC$ circumscribes a circle, with points of contact being $X,$$Y$$,Z$. If the feet of the altitudes of $\Delta XYZ$ are $D,$$E$,$F$ , prove that the sides of $\Delta DEF$ are parallel to the sides of $\Delta ABC$.
This is an olympiad book question related to tangents and its theorems.
I tried to solve it by using alternate segment theorems and other theorems which would help solving this but I fail to prove the statement.
In the figure, H, is the ortho-center of XYZ. Then, $\angle 6 = \angle 7$.
I is the in-center of ABC. Then, all the red-dotted angles are equal.
By angles in alternate segment, the green marked angles are equal.
$\angle 6 = \angle 7 = 90^0 -\angle XZY = 90^0 – \angle XYC = \angle 1 =\angle 3$. Result follows.