Let $ABC$ be a triangle inscribed in a circle, and $P$ be a point on this circle. Let $P \in AB$, $N \in AC$ and $R \in BC$ be such that $PM\perp AB$, $PN\perp AC$ and $PR\perp BC$. Prove that $MNR$ is a line, called the Simson line of $P$.
My attempt, $\angle ANP=90$
$ANMP$ is a cyclic quadrilateral.
What should I do further.?
On
Initially, we have a point P lying on the circle circumscribing $\triangle ABC$. From P. perpendicular lines $(\alpha , \beta , \gamma )$ are drawn to AB, BC, CA respectively at M, R and N.
The result is MNR forms a straight line and is called the Simson’s line.
The proof goes like this:-
Initially, MN and NR are two different straight lines.
From those right angles, we have the red and green dotted circles.
From them, we get $\angle 1 = \angle 2$, $\angle 3 = \angle4$, and $\angle 1 = \angle 4$.
The fact that $\angle 2 = \angle 4$ forces MN and NR to form a straight line.
Hint: It suffices to prove that $\angle MNA = \angle CNR$. Why? Because that will easily imply that $\angle MNR = \angle ANC = \pi.$
Further hint: $\angle MNA = \angle MPA = \frac{\pi}{2} - \angle MAP = \frac{\pi}{2} - (\pi - \angle BAP) = \angle BAP - \frac{\pi}{2}$. Now use the fact that $MNRC$ and $ABPC$ are cyclic quadrilaterals in much the same way.