We could just draw a line and try to make it perpendicular, but then we have no guarantee that it's perfectly perpendicular.

How can you guarantee that a line is perpendicular?

If we pick two points on the perpendicular line which are an equal distance from the intersection, they will also be the same distance from every other point on the line we started with.

If we don't already have the perpendicular line, is there another way to find the blue points?

If we use the compass to put a circle somewhere on the line, the circle will include all points that are the same distance from that point, including the two blue points.

We can add a second circle somewhere else on the line that intersects with the first circle.

The points where the two circles intersect can be used to draw a perpendicular line.

Use a straightedge to connect the two points where the circles intersect. This line is perpendicular to the given line.

We could just draw a line at P and try to make it parallel, but then we have no guarantee that it's perfectly parallel.

If we did have parallel line and drew a line from a point A on the given line, through P, then the angle at A must be the same as the angle at P.

Therefore, if we copy the angle at A to point P, we will be able to construct parallel lines.

If we construct a triangle with a vertex at point A, then construct a triangle with the same side lengths at and a vertex at point P, the angles at A and P will be the same.

We can use a compass centered at A to find all the point a given distance from A.

We can use a compass with same radius centered at P to find all the points the same distance from P. We can ensure we have get the same radius by putting the compass at point A first, getting the radii equal, then moving the compass to point P.

To find the distance between the two points equidistant from A we can add center a compass on one point and change the radius to go through the other point.

Now move the compass to the point where the first compass intersected with the line through P.

Now we can use the point where the two compasses intersect to construct a triangle with the same side lengths as the triange at point A.

Finally use a straightedge to connect point P to where the two compasses intersect. This line will be parallel to the original line.

We could just draw a line and try to get it right, but then we have no guarantee that it's actually perfectly perpendicular or that it bisects the segment at exactly the midpoint.

How can you guarantee that a line is really a perpendicular bisector?

If we pick two points on the perpendicular bisector which are an equal distance from the intersection, they will also be the same distance from both endpoints of the segment we started with.

If we don't already have the perpendicular bisector, is there another way to find the blue points?

If we use the compass to put a circle centered at point A, the circle will include all points that are the same distance from point A, including the two blue points.

We can add a second circle at point B that intersects with the first circle.

But wait! We can use these circles to draw a perpendicular line, but not a bisector! That's because the two circles are different sizes.

One nice way to make the circles the same size is to set the radii equal to the distance between A and B. You can do this by setting the center at one point and the edge of the circle at the other.

Use a straightedge to connect the two points where the circles intersect. This line segment is the perpendicular bisector of \overline{AB}.

We could just draw a line and try to make it bisect the angle, but that's difficult to do and there is no guarantee it's a perfect bisector.

If we pick any two points on the given lines that are the same distance from the vertex of the angle, every point on the bisector line will be equidistant from those points.

We can use a compass centered at the vertex to find two points equidistant from the vertex.

If we use two more compasses centered at each of the two points, we can see that they intersect, but not on the angle bisector! That's because the two circles are not the same size.

A nice way to make the circles the same size is to set the edges of both circles so they pass through the vertex.

Use a straightedge to connect the vertex to the point where the last two circles intersect. This line is the angle bisector.

We could just draw a line at P and try to make it the same length as \overline{AB}, but then we have no guarantee that it's actually the same length.

How could we find all the points the correct length from P?

If we center a compass on A and set its radius such that it intersects B, then we will have a compass with a radius of the correct length.

Now we can move the compass to P.

Finally use a straightedge to connect point P to any point on the compass circumference.

First we need to draw a line through P.

We could just draw a second line through P and try to make the angle the same as \angle BAC, but that's difficult to do and there is no guarantee the angle will be exactly the same.

If we construct a triangle with a vertex at point A, then construct a triangle with the same side lengths and a vertex at point P, the angles at A and P will be the same.

We can use a compass centered at A to find two points equidistant from A.

We can use a compass with same radius centered at P to find all the points the same distance from P. We can ensure we get the same radius by putting the compass at point A first, getting the radii equal, then moving the compass to point P.

To find the distance between the two points equidistant from A we can center a compass on one point and change the radius to go through the other point.

Now move the compass to the point where the first compass intersected with the line through P.

Finally use a straightedge to connect point P to where the two compasses intersect. The angle formed at point P will be the same as \angle BAC.