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$ABC$ is an inscribed triangle, $PM\perp AB$, $PN\perp AC$ and $PR\perp BC$. Prove that MNR is called Simson line.

My attempt, $\angle ANP=90$

$ANMP$ is a cyclic quadrilateral.

What should I do further.?

What is Simpson line? I don't know it's meaning as wellenter image description here

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    How do you prove a definition? You can show that $M$, $N$, and $R$ are collinear, but it is the definition that they form the Simson line of $P$ with respect to $A$, $B$, and $C$. (It's Simson, not Simpson, by the way.) – 2017-01-28
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    Here's Simpsons' line explained: https://www.youtube.com/watch?v=Q9A0Vufw3NQ – 2017-01-28

2 Answers 2

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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.

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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.

enter image description here

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.