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Let A B C D be any four points. Then the angles angle ABC + angle CDA = if and only if the four points lie on a circle. As a corollary, you may conclude that if the angles of the triangle ABC at the vertices A B and C are alpha, beta and gamma respectively, then the angle ADC is equal to pi - beta if and only if D lies on the circle.

Need to prove that X,Y,Z are collinear iff angle ADC = pi - beta. X, Y, Z are points of intersection of perpendiculars from point D to the sides of triangle ABC

I can see when D lies on the circle and form a quadrilateral ABCD, AC will be the diagonal of that quadrilateral. It will also lead to forming two triangles ABC and ADC. Other than that, I'm not really sure how to proceed.

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I wanted to give you some hints, but it soon got too complicated, so I'm posting the full solution. There are other approaches, but I like the one using reflections the most: I really find it simpler than the one at Wikipedia page, and also it is my own invention (reinvention probably, though).

I'm assuming that $X$, $Y$, and $Z$ are the orthogonal projections of $D$ onto $AB$, $BC$ and $CA$. If so, then you are trying to prove the existence of Simson line. Consider the following picture:

$\hspace{20pt}$simpson

where $AA'$ is the diameter and $D_{AB}$, $D_{BC}$, $D_{CA}$ are reflections of $D$ across $AB$, $BC$ and $CA$ respectively. From properties of reflection (composition of two reflections is a rotation around the intersection of their axes) we know that $D_{AB}$ is an image of $D_{BC}$ in rotation around $B$ by angle $2\angle CBA$ (i.e. the triangle $\triangle D_{AB}BD_{BC}$ is isosceles), hence $$\angle D_{AB}D_{BC}B = 90^\circ-\angle ABC = \angle A'BC.$$ For similar reasons, $\angle D_{CA}D_{BC}C = \angle BCA$. Moreover, since $D_{BC}$ is a reflection of $D$ via $BC$, we have $\angle BD_{BC}C = \angle BDC$. Obviously $D_{AB}$, $D_{BC}$ and $D_{CA}$ are collinear if and only if green and blue angles sum up to $180^\circ$, in other words $\angle BDC = \angle BA'C$, but that happens if and only if $P$ belongs to the circumcircle of $ABC$.

Finally, points $X$, $Y$ and $Z$ are images of $D_{AB}$, $D_{BC}$ and $D_{CA}$ respectively in homothety centered at $D$ and ratio $\frac{1}{2}$ (that is, for example, $X$ is the midpoint of $DD_{AB}$). It follows that $X$, $Y$ and $Z$ are collinear if and only if $D_{AB}$, $D_{BC}$ and $D_{CA}$ are collinear, that is, if and only if $D$ belongs to circumcircle of $ABC$.

It is worth noting, that if $D_{AB}$, $D_{BC}$ and $D_{CA}$ are collinear, then they are also collinear with the orthocenter of $ABC$, e.g. see here. Check out also the Wikipedia and MathWorld.

I hope this helps ;-)

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Awesome!!! Thank you –  devcoder Apr 4 '13 at 1:06

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