Constructing Perpendiculars

Construction of Perpendicular to a Line from a Point on the Line

A perpendicular is a line or ray that intersects another line or line segment at a $90^\circ$ angle. Constructing a perpendicular from a point *on* the line is equivalent to constructing a $90^\circ$ angle at that specific point on the line.

Construction Steps

Given: A line $l$ and a point $P$ that lies on the line $l$.

Tools: Compass, Straightedge.

Goal: Construct a line (or ray) passing through $P$ that is perpendicular to line $l$.

Steps:

1. Mark Equidistant Points on the Line: Place the pointed end of the compass on point $P$. Using any convenient compass radius, draw an arc that intersects the line $l$ on both sides of $P$. Let the points of intersection be $A$ and $B$. (Note: By this step, the segment $AB$ is centered at $P$, so $AP = BP$).
2. Draw Intersecting Arcs (Above or Below the Line): Place the compass point on $A$. Choose a new compass radius that is greater than the distance $AP$ (or $BP$). Draw an arc on one side of the line $l$ (either above or below).
3. Draw Second Intersecting Arc: Keeping the same compass width that was used in Step 2, place the pointed end of the compass on point $B$. Draw another arc that intersects the arc drawn from $A$. Label the point of intersection of these two arcs as $Q$.
4. Draw the Perpendicular: Use the straightedge to draw a straight line passing through the given point $P$ and the point of intersection $Q$. Extend this line as needed.

The line $PQ$ is perpendicular to the line $l$ at point $P$. This means that the angles formed at $P$, such as $\angle QPA$ and $\angle QPB$, are both $90^\circ$.

Justification:

This construction is essentially the same as constructing a $90^\circ$ angle at a point on a line, or constructing the perpendicular bisector of the segment $AB$ where $P$ is the midpoint of $AB$. As shown in the justification for the $90^\circ$ angle construction (I1) and the perpendicular bisector construction (I4), the line $PQ$ is the perpendicular bisector of segment $AB$. Since $P$ lies on the line segment $AB$ (and is its midpoint by construction), the perpendicular bisector $PQ$ passes through $P$ and is perpendicular to $AB$ (which lies on line $l$). Thus, $PQ \perp l$ at $P$.

Example

Answer:

Construction of Perpendicular to a Line from a Point outside the Line

This construction involves "dropping" a perpendicular from a point that is not on the given line to the line. The resulting segment from the point to the line, along the perpendicular, represents the shortest distance from the point to the line.

Construction Steps

Given: A line $l$ and a point $P$ that is not on line $l$.

Tools: Compass, Straightedge.

Goal: Construct a line passing through $P$ that is perpendicular to line $l$. Let this line intersect $l$ at $M$. The segment $PM$ will be perpendicular to $l$.

Steps:

1. Draw Arc Intersecting the Line: Place the pointed end of the compass on the point $P$ (outside the line). Choose a compass radius that is large enough so that when you draw an arc, it intersects the line $l$ at two distinct points. Draw such an arc. Label the points where this arc intersects line $l$ as $A$ and $B$. (Note: By this step, $PA = PB$ as they are radii of the same arc).
2. Draw Intersecting Arcs (on the Opposite Side): Now, with $A$ as the center and using any convenient compass radius (it's common to use a radius greater than half of $AB$, or even the same radius as Step 1, provided the arcs intersect), draw an arc on the side of line $l$ opposite to where point $P$ is located.
3. Draw Second Intersecting Arc: Keeping the same compass width that was used in Step 2, place the pointed end of the compass on point $B$. Draw another arc that intersects the arc drawn from $A$ in Step 2. Label the point of intersection of these two arcs as $Q$.
4. Draw the Perpendicular: Use the straightedge to draw a straight line passing through the given point $P$ and the point of intersection $Q$. Let this line $PQ$ intersect the original line $l$ at point $M$.

The line $PQ$ is the perpendicular line from $P$ to $l$. The line segment $PM$ is the required perpendicular from $P$ to $l$, and $PM \perp l$.

Justification:

Consider points $P$, $A$, $B$, and $Q$. By construction, $PA = PB$ (radii from $P$) and $QA = QB$ (arcs drawn with the same radius from $A$ and $B$). A point that is equidistant from the endpoints of a line segment lies on the perpendicular bisector of that segment. Since both $P$ and $Q$ are equidistant from $A$ and $B$, the line passing through $P$ and $Q$ must be the perpendicular bisector of the line segment $AB$. Since the line segment $AB$ lies on the line $l$, the perpendicular bisector of $AB$ is perpendicular to the line $l$. Therefore, the line $PQ$ is perpendicular to line $l$. As $M$ is the point where $PQ$ intersects $l$, the segment $PM$ is the perpendicular from $P$ to $l$.

Example

Answer:

Construction of Perpendicular Bisector of a Line Segment

The perpendicular bisector of a line segment is a line that fulfills two conditions simultaneously: it is perpendicular to the segment, and it passes through the segment's midpoint. Any point on the perpendicular bisector is equidistant from the two endpoints of the segment.

Construction Steps

Given: A line segment $AB$.

Tools: Compass, Straightedge.

Goal: Construct the line that is perpendicular to $AB$ and passes through its midpoint.

Steps:

1. Draw Arcs from Endpoint A: Place the pointed end of the compass on endpoint $A$. Choose a compass radius that is more than half the length of the segment $AB$. This is important to ensure the arcs intersect. Draw an arc above the line segment $AB$ and another arc below the line segment $AB$.
2. Draw Arcs from Endpoint B: Keeping the same compass width that was used in Step 1, place the pointed end of the compass on endpoint $B$. Draw an arc above the line segment $AB$ and another arc below the line segment $AB$. These arcs should intersect the corresponding arcs drawn from point $A$. Label the point of intersection above $AB$ as $P$ and the point of intersection below $AB$ as $Q$. (Note: By this step, $AP = BP$ and $AQ = BQ$, as these are radii of arcs drawn with the same compass settings from $A$ and $B$).
3. Draw the Perpendicular Bisector: Use the straightedge to draw a straight line passing through the two intersection points $P$ and $Q$.

The line $PQ$ is the perpendicular bisector of the line segment $AB$. If $M$ is the point where the line $PQ$ intersects the line segment $AB$, then $M$ is the midpoint of $AB$, and the line $PQ$ is perpendicular to $AB$ at $M$.

Example

Answer:

Justification of Perpendicular Bisector Construction

The justification proves that the line $PQ$, constructed as the perpendicular bisector of segment $AB$ (as in I3), is indeed perpendicular to $AB$ and passes through its midpoint.

Proof using Congruent Triangles

Given: Line segment $AB$. Line $PQ$ is constructed such that $P$ and $Q$ are points on opposite sides of $AB$ where arcs of equal radius from $A$ and $B$ intersect. Let $M$ be the point where $PQ$ intersects $AB$.

To Prove:

1. $PQ \perp AB$ (The line $PQ$ is perpendicular to $AB$)
2. $AM = BM$ ($M$ is the midpoint of $AB$)

Construction for Proof: Join points $A$ and $B$ to the intersection points $P$ and $Q$. This forms four line segments: $AP, BP, AQ, BQ$. Let $M$ be the intersection of $PQ$ and $AB$.

Proof:

Part 1: Proving $\triangle PAQ \cong \triangle PBQ$

Part 2: Proving $\triangle APM \cong \triangle BPM$

Now consider the segments $AP$ and $BP$ and the line $PQ$ intersecting $AB$ at $M$.

Part 3: Proving Perpendicularity

The angles $\angle AMP$ and $\angle BMP$ are adjacent angles formed on the straight line $AB$ where the ray $PM$ meets it. Therefore, they form a linear pair.

From Part 2, we know that $\angle AMP = \angle BMP$. Substitute this into the equation:

Since $\angle AMP = 90^\circ$, the line $PQ$ is perpendicular to the line segment $AB$ at $M$.

Conclusion: We have proved that $AM = BM$ (PQ bisects AB) and $\angle AMP = 90^\circ$ (PQ is perpendicular to AB). Therefore, the line $PQ$ constructed is the perpendicular bisector of the line segment $AB$. Q.E.D.