# How to prove this three-dimensional geometry theorem?

Given that point $P$ outside plane $\alpha$. Plane $\beta$ is a plane that contained point $P$ and perpendicular to one of the line in plane $\alpha$. If line $l$ is the intersection line between plane $\alpha$ and plane $\beta$, then $d(P,\alpha) = d(P,l)$

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A terminological comment: one does not prove problems. One may prove a theorem, or solve a problem. – Gerry Myerson Feb 20 '13 at 0:08
edited.., thanks... :) – chihiroasleaf Feb 20 '13 at 0:12

Consider the point $R$ on $\alpha$ minimizing $d(P,R)$. By definition, this is $d(P,\alpha)$, and a basic fact about distances between points and planes tells us that the segment $PR$ will be perpendicular to $\alpha$. Moreover, $R$ is unique, and so is the segment $PR$
The plane $\beta$ is uniquely defined by the point $P$ and the line $l$ where it intersects $\alpha$. There is a point $L$ on $l$ minimizing the distance $d(P,L)$. $L$ is unique, and it is the only point such that the segment $PL$ is perpendicular to $L$.
Because $\alpha$ and $\beta$ are perpendicular, any segment in $\beta$ perpendicular to $l$ (their intersection) must be perpendicular to alpha. Therefore $PL$ is perpendicular to $\alpha$.
This is enough to deduce that $PL$ and $PR$ are on the same line. Because $L$ and $R$ are both in $\alpha$, they must be the same. Therefore the lengths of $PL$ and $PR$ are equal, and by definition $d(P,\alpha) = d(P,l)$.