Practical Geometry - Constructing Perpendiculars

Construct a perpendicular bisector of a line segment $XY$ of length $8 \text{ cm}$ using a ruler and compasses.

1. Draw a line segment $XY = 8 \text{ cm}$ using a ruler.
2. With $X$ as center and a radius more than $4 \text{ cm}$ (half of $XY$), draw two arcs, one above and one below $XY$.
3. With $Y$ as center and the same radius, draw two arcs cutting the previous arcs at points $P$ and $Q$.
4. Join $P$ and $Q$.
5. The line $PQ$ intersects $XY$ at point $M$. Here, $PQ \perp XY$ and $XM = MY = 4 \text{ cm}$.

To bisect a segment, the compass radius must be greater than $\frac{1}{2}$ the length (i.e., $> \frac{1}{2} \times 8 = 4 \text{ cm}$) so that the arcs from both ends can actually intersect.

Draw a line $l$ and a point $A$ on it. Construct a perpendicular to $l$ through $A$ using compasses.

1. Draw a line $l$ and mark a point $A$ on it.
2. With $A$ as center and any convenient radius, draw an arc that cuts the line $l$ at two points, $B$ and $C$.
3. With $B$ as center and a radius greater than $AB$, draw an arc above the line.
4. With $C$ as center and the same radius as in step 3, draw another arc cutting the previous arc at point $D$.
5. Join $AD$. The line $AD$ is the required perpendicular to line $l$ at point $A$.

By creating points $B$ and $C$ equidistant from $A$, we ensure $A$ is the midpoint. Any point $D$ equidistant from $B$ and $C$ must lie on the perpendicular passing through the midpoint $A$.

1. Draw a line segment $PQ = 6 \text{ cm}$.
2. Mark a point $R$ on $PQ$.
3. With $R$ as center and a convenient radius, draw an arc intersecting $PQ$ at two points, $X$ and $Y$.
4. With $X$ as center and radius greater than $RX$, draw an arc above the line.
5. With $Y$ as center and the same radius, draw another arc intersecting the previous arc at point $S$.
6. Join $RS$. $RS$ is the required perpendicular to $PQ$.

This method uses the property that the perpendicular to a line at a point is the locus of points equidistant from two points on the line that are equidistant from the given point $R$.

1. Draw line $m$ and mark point $A$ outside it.
2. With $A$ as center and a radius large enough to intersect $m$ at two points, draw an arc cutting $m$ at $B$ and $C$.
3. Using $B$ and $C$ as centers and a radius more than half of $BC$, draw two arcs on the opposite side of $A$ that intersect at point $D$.
4. Join $AD$. Line $n$ (line $AD$) is perpendicular to line $m$.

Since $A$ and $D$ are both equidistant from $B$ and $C$, the line joining them must be the perpendicular bisector of the segment $BC$, which means $AD$ is perpendicular to line $m$.