Proposition IV.8 — In a given square to inscribe a circle.

Proof

Let

A B C D

be the given square. Bisect the sides at

E

F

G

H

(I.10). Join

E G

and

F H

, meeting at

K

. By I.34 and the construction,

K E = K F = K G = K H

. Drop perpendiculars from

K

to each side; each is equal to

K E

. Thus the circle with centre

K

and radius

K E

touches each side at its midpoint (III.16).

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Depends on (3)

I.10Proposition I.10To bisect a given finite straight line.
I.34Proposition I.34In parallelogrammic areas the opposite sides and angles are equal to one another, and the diameter bisects the areas.
III.16Proposition III.16The straight line drawn at right angles to the diameter of a circle from its extremity will fall outside the circle,…

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