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Exercise 205: *Between the sides of a given angle, place a segment congruent to a given segment and such that it cuts congruent segments on the sides of the angle.*

Let the given angle be $\angle ABC$ and the given segment be $MN$.

Use the method in $\S 64$ to bisect $\angle ABC$, the bisector is $BD$.

Use the method in $\S 67$ to bisect the segment $MN$, midpoint $O$.

Use the method in $\S 65$ to erect a perpendicular to $BD$ at $B$, extend it into the plane on both sides of $BD$.

Use a compass to mark a segment congruent to $MO$ originating at $O$ and in the direction of the plane containing $A$, terminal point $P$.

Make another segment of the same originating at $O$, but in the opposite direction, terminal point $Q$.

Erect a perpendicular at $P$ in the direction of $D$, intersection with side $AB$ is $R$.

Erect a perpendicular at $Q$ in the direction of $D$, intersection with $CB$ is $S$. Connect $R$ and $S$ with a line.

$PR \parallel QS$, $\S 71$) two perpendiculars to the same line are parallel.

$\angle PBR = d - \angle ABD$; $\angle ABD = \angle CBD$; $\angle QBS = d - \angle CBD$; all by construction. $\angle PBR = \angle QBS$.

$\angle BPR = d = \angle BQS$, by construction. $BP = BQ$, by construction. $\triangle BPR = \triangle BQS$, $\S 40(2)$ ASA.

$BR = BS$ and $PR = QS$. $QPRS$ is a parallelogram, $\S 86(2)$ two opposite congruent and parallel sides.

$MN = PQ = RS$, $\S 85$) parallelogram opposite sides are congruent.

$RS$ is the required segment between sides, and $BR$ and $BS$ are the congruent cut-off segments along each side.