The English works of Thomas Hobbes of Malmesbury, Volume 01 (of 11)
Thomas Hobbes · en
[Sidenote: If two strait lines drawn from the same point without a circle fall upon the circumference, and the lines reflected from them being produced meet within
the circle, they will make an angle equal to twice that angle, which is made by two strait lines drawn from the centre to the points of incidence, together with the angle which the incident lines themselves make.]
4. If two strait lines drawn from the same point without a circle fall
upon the circumference, and the lines reflected from them being produced
meet within the circle, they will make an angle equal to twice that
angle, which is made by two strait lines drawn from the centre to the
points of incidence, together with the angle which the incident lines
themselves make.
Let the two strait lines A B and A C (in fig. 4) be drawn from the point
A to the circumference of the circle, whose centre is D; and let the
lines reflected from them be B E and C G, and, being produced, make
within the circle the angle H; also let the two strait lines D B and D C
be drawn from the centre D to the points of incidence B and C. I say,
the angle H is equal to twice the angle at D together with the angle at
A.
For let A C be produced howsoever to I. Therefore the angle I C H, which
is external to the triangle C K H, will be equal to the two angles C K H
and C H K. Again, the angle I C D, which is external to the triangle C L
D, will be equal to the two angles C L D and C D L. But the angle I C H
is double to the angle I C D, and is therefore equal to the angles C L D
and C D L twice taken. Wherefore the angles C K H and C H K are equal to
the angles C L D and C D L twice taken. But the angle C L D, being
external to the triangle A L B, is equal to the two angles L A B and L B
A; and consequently C L D twice taken is equal to L A B and L B A twice
taken. Wherefore C K H and C H K are equal to the angle C D L together
with L A B and L B A twice taken. Also the angle C K H is equal to the
angle L A B once and A B K, that is, L B A twice taken. Wherefore the
angle C H K is equal to the remaining angle C D L, that is, to the angle
at D, twice taken, and the angle L A B, that is, the angle at A, once
taken; which was to be proved.