Proposition IV.4 — In a given triangle to inscribe a circle.

Proof

Let

A B C

be the given triangle. Bisect

∠ A B C

and

∠ B C A

B D

and

C D

(I.9), meeting at

D

. Drop perpendiculars

D E

D F

D G

from

D

A B

B C

C A

(I.12). In the pairs of triangles formed at

D

, the two angles and a common side give congruence (I.26), whence

D E = D F = D G

. The circle with centre

D

and radius

D E

touches each side (since the perpendiculars at the feet make the sides tangents by III.16) and is inscribed in

△ A B C

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Full neighborhood

Depends on (4)

I.9Proposition I.9To bisect a given rectilineal angle.
I.12Proposition I.12To draw a perpendicular straight line to a given infinite straight line from a given point not on it.
I.26Proposition I.26If two triangles have the two angles equal to two angles respectively, and one side equal to one side, namely either…
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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