Proposition IV.5 — About a given triangle to circumscribe a circle.

Proof

Let

A B C

be the given triangle. Bisect

A B

D

and

A C

E

(I.10). From

D

and

E

draw perpendiculars to

A B

and

A C

respectively (I.11), meeting at

F

. Join

F A

F B

F C

. By I.4 applied to the two right triangles at

D

F A = F B

; similarly at

E

F A = F C

. Thus

F A = F B = F C

, and the circle with centre

F

and radius

F A

passes through all three vertices.

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Full neighborhood

Depends on (4)

I.4Proposition I.4If two triangles have two sides equal to two sides respectively, and have the angles contained by the equal straight…
I.10Proposition I.10To bisect a given finite straight line.
I.11Proposition I.11To draw a straight line at right angles to a given straight line from a given point on it.
I.15Definition I.15A circle is a plane figure contained by one line such that all the straight lines falling upon it from one point among…

Required by (dependents) (1)

IV.10Proposition IV.10To construct an isosceles triangle having each of the angles at the base double of the remaining one.

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