Given AB | Construct perpendicular to AB at B | Prop 11 | Construct circle with center B and radius AB | Post 3 |
Extend AB | Post 2 |
Let Z be the 2nd intersection of ray AB and the circle | Def 17 |
Construct circles with centers A and Z and radii AZ | Post 3 |
Let X and Y be their intersections | Assumption that 2 circles in a plane intersect in 2 pts |
Construct XY | Post 1 |
Notice that AX=XZ. | They are radii of circles with radius AZ, Def 15, and CN 1. |
Notice that XB=XB. | CN4 |
Notice that AB=BZ. | Def 15 as they are radii of the circle with center B and radius AB. |
Since triangle ABX is congruent to triangle ZBX, then angle XBA= angle XBZ. | Prop 8 (SSS) |
Hence angles XBA and XBZ are right. | Def 10 |
Construct C as the intersection of XB and the circle at B of radius AB. | Def 15 |
Notice that BC=AB. | Def 15 |
In addition, angle ABC is right | Angle ABA is right (above), C is on XB, Def 8, CN4. |
.... | |
We have constructed a figure with 4 equal sides and 4 right angles. Thus we have constructed a square. | Def 22 |
square2