- Ellie Krumsieg
Proposition 4.9 - About a given square, circumscribe a circle.
Proof Let ABCD be a given square such that AB=BC=CD=AD. Let AC and BD be made such that E is the intersection of AC and BD. Now, consider triangles ABC and ADC. AD=AB, AC=AC, and BC=CD. So, by SSS, ABC is congruent to ADC. Thus, <BAC=<CAD which implies that AC bisects <BAD. Similarily, <ADC,<DCB, and <ABC are bisected by BD,AC, and DB respectively. Since, each angle of the square ABCD is equal, the angles formed by the bisections are equal. For example, <EAB=<EBA because <DAB=<ABC and <EAB bisects <DAB just as <EBA bisects <ABC. Therefore, by isosceles triangles, EA=BE because triangle ABE's base angles are congruent. Similarily, EA=ED=EC=EB. Therefore, it can be said that the circle with center E and radius EA is circumscribed about the square ABCD.