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The following exercise is from the book What is Mathematics by Richard Courant and Herbert Robbins:

Given two lines L, M and two points P, Q situated inside the angle formed by the two lines, the path of minimum length from P to L, then to M, and then to Q can be found as follows:

Let Q' be the reflection of Q in M and Q'' the reflection of Q' in L; draw PQ'' intersecting L in R and RQ' intersecting M in S; then R and S are the required points such that PR+RS+SQ is the path of minimum length from P to L to M to Q.
One might ask for the shortest path first from P to M, then to L, and from there to Q. In this case Q' is the reflection of Q in L and Q'' the reflection of Q' in M. R is the intersection of PQ'' with M and S the intersection of RQ' with L.

one image second image

Show that the first path is smaller than the second if O (where O is the point of intersection between L and M) and R lie on the same side of the line PQ.

How do you do this?

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The diagrams above depicts straight lines obtaining by developing 3D helical geodesics on a cone of semi-vertical angle $ π\cdot \sin^{−1}LOM.$ Generators OL,OM are diametrically opposite lines of the cone seen when a thin flexible transparent sheet cone is flattened to a plane,when the geodesic curved helices now appear as straight lines by virtue of Gauss Egregium Theorem (in this particular case straight lines map to straight lines). The global minimum geodesic is the line PQ. There are infinitely many geodesics /shortest lines of course. – Narasimham Aug 26 at 8:58

A practical method is making 3 cone sector developments and joining P to Q cutting the generator borders.

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same comment re-posted as answer after a month. – Narasimham Sep 26 '15 at 18:48

If Q'' is the final point obtained by first reflecting Q across M and to obtain Q' and then reflecting Q' across L to obtain Q'', and P'' is the final point obtained by first reflecting Q across L to obtain P' and then reflecting P' across M to obtain P'', then the length of segment PQ'' is the length of the first path and the length of the segment PP'' is the length of the second path. If you show that the line through O and Q is the perpendicular bisector of the segment joining Q'' and P'' the problem is solved. To do this first notice that Q, Q', P', Q'' and P'' lie on the circle with center O passing through Q (this follows easily from the construction), and then that angle QQ'Q'' equals angle QP'P'' so that the arcs they intercept are equal. Also the radii OQ'' and OP'' are equal. From this you can prove that OQ is the perpendicular bisector of the segment joining Q'' and P''

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An alternative approach:

Construct $\angle KOL$ "clockwise" of $\angle LOM$ and likewise $\angle JOK$ "clockwise" of $\angle KOL$ so that $m\angle JOK = m\angle KOL = m\angle LOM$. Then add images $P^*$ and $Q^*$ of $P$ and $Q$ so that $P^*$ and $Q^*$ are arranged the same way in $\angle JOK$ as $P$ and $Q$ are in $\angle LOM$.

The two path lengths in question are equal to $PQ^*$ and $P^*Q$. Use the law of cosines on $\triangle POQ^*$ and $\triangle P^*OQ$, along with the fact that cosine is monotonically decreasing on $(0, \pi)$, and the proposition should follow.

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