proof of Simson’s line

Given a $\mathrm{△}ABC$ with a point $P$ on its circumcircle (other than $A,B,C$),we will prove that the feet of the perpendiculars drawn from P to the sides $AB,BC,CA$(or their prolongations) are collinear.

Since $PW$ is perpendicular to $BW$ and $PU$ is perpendicular to $BU$the point $P$ lies on the circumcircle of $\mathrm{△}BUW$.

By similar arguments, $P$ also lies on the circumcircle of $\mathrm{△}AWV$ and $\mathrm{△}CUV$.

This implies that $PUBW$ , $PUCV$ and $PVWA$ are all cyclic quadrilaterals.

Since $PUBW$ is a cyclic quadrilateral,

$$\mathrm{\angle }UPW={180}^{\circ }-\mathrm{\angle }UBW$$

implies

$$\mathrm{\angle }UPW={180}^{\circ }-\mathrm{\angle }CBA$$

Also $CPAB$ is a cyclic quadrilateral, therefore

$$\mathrm{\angle }CPA={180}^{\circ }-\mathrm{\angle }CBA$$

(opposite angles in a cyclic quarilateral are supplementary).

From these two, we get

$$\mathrm{\angle }UPW=\mathrm{\angle }CPA$$

Subracting $\mathrm{\angle }CPW$, we have

$$\mathrm{\angle }UPC=\mathrm{\angle }WPA$$

Now, since $PVWA$ is a cyclic quadrilateral,

$$\mathrm{\angle }WPA=\mathrm{\angle }WVA$$

also, since $UPVC$ is a cyclic quadrilateral,

$$\mathrm{\angle }UPC=\mathrm{\angle }UVC$$

Combining these two results with the previous one, we have

$$\mathrm{\angle }WVA=\mathrm{\angle }UVC$$

This implies that the points $U,V,W$ are collinear.