III.11 Proposition III.11
If two circles touch one another internally, and their centres be taken, the straight line joining their centres, if produced, will fall on the point of contact of the circles.
Proof
Let circle Γ1 contain circle Γ2, touching at A,
with centres F (of Γ1) and G (of Γ2). Suppose
the line FG produced does not pass through A. Join FA, GA.
In △FAG: by the triangle inequality (I.20), FA+AG>FG. Produce FG to meet Γ1 at H and Γ2 at K.
Then FA=FH (radii of Γ1), GA=GK (radii of
Γ2), and H lies beyond K on segment FG extended. So
FA+AG=FH+GK=FG+(HK>0), i.e.\ FA+AG>FG —
consistent. But Γ2 is internally tangent, so H=K, and
FA+AG=FG, contradicting the strict inequality. Hence A
lies on line FG.
lines 74–74 in main.tex
