Proposition III.11 — If two circles touch one another internally, and their centr…

Proof

Let circle

Γ_{1}

contain circle

Γ_{2}

, touching at

A

, with centres

F

(of

Γ_{1}

) and

G

(of

Γ_{2}

). Suppose the line

F G

produced does not pass through

A

. Join

F A

G A

. In

△ F A G

: by the triangle inequality (I.20),

F A + A G > F G

. Produce

F G

to meet

Γ_{1}

H

and

Γ_{2}

K

. Then

F A = F H

(radii of

Γ_{1}

G A = G K

(radii of

Γ_{2}

), and

H

lies beyond

K

on segment

F G

extended. So

F A + A G = F H + G K = F G + (H K > 0)

, i.e.\

F A + A G > F G

— consistent. But

Γ_{2}

is internally tangent, so

H = K

, and

F A + A G = F G

, contradicting the strict inequality. Hence

A

lies on line

F G

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Full neighborhood

Depends on (3)

I.20Proposition I.20In any triangle two sides taken together in any manner are greater than the remaining one.
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…
III.3Definition III.3Circles are said to touch one another which, meeting one another, do not cut one another.

Required by (dependents) (2)

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