# proof of Simson’s line

Given a $\triangle ABC$ 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 $\triangle BUW$.

By similar arguments, $P$ also lies on the circumcircle of $\triangle AWV$ and $\triangle CUV$.

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

Since $PUBW$ is a cyclic quadrilateral,

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

implies

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

Also $CPAB$ is a cyclic quadrilateral, therefore

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

(opposite angles in a cyclic quarilateral are supplementary).

From these two, we get

 $\angle UPW=\angle CPA$

Subracting $\angle CPW$, we have

 $\angle UPC=\angle WPA$

Now, since $PVWA$ is a cyclic quadrilateral,

 $\angle WPA=\angle WVA$

also, since $UPVC$ is a cyclic quadrilateral,

 $\angle UPC=\angle UVC$

Combining these two results with the previous one, we have

 $\angle WVA=\angle UVC$

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

Title proof of Simson’s line ProofOfSimsonsLine 2013-03-22 13:08:26 2013-03-22 13:08:26 giri (919) giri (919) 9 giri (919) Proof msc 51-00