Proposition 12
To a given infinite straight line, from a given point which is not on it, to draw a perpendicular straight line.
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Before we begin, here is the definition for right angles and perpendicular lines:
When a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right, and the straight line standing on the other is called a perpendicular to that on which it stands.To show that two lines are perpendicular we must show that the two lines form right angles.
Here is an infinite straight line AB and a point C that is not on it. The goal is to draw a line from point C that is perpendicular to AB.
Let point D be any point on the other side of AB from C, and draw a circle whose radius is equal to CD. I have labeled the circle as EFG, where points E and G are the points where the circle intersects the line AB.
Let the point H be the point at which the segment EG is bisected. (Proposition 10 covered how to find the point that bisected a line segment.)
Draw the segments from point C to the points G, H, and E.
We will now focus on triangles GCH and ECH as we try to show that the segment CH is perpendicular to AB.
- GH is equal to HE, since H bisected EG.
- CH is common to both triangles.
- CG is equal to CE, because they are both on the circle with center C and radius CD.
So, by Proposition 8 (if two triangles have two equal sides and equal bases, then the angles between the equal sides are equal to each other, or in today’s introductory geometry classes: side-side-side implies equal angles), the two angles GHC and EHC are equal to each other.
Since the line segment CH set up on the straight line AB makes two angles GHC and EHC that are adjacent angles and equal to each other, both angles GHC and EHC are right angles.
By definition, the line segment CH is perpendicular to the line AB.
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