212. A Demonſtration of the Inclination of the Looking-
Glaſs.

48. Let A B be a Ray, proceeding from ſome
Point of an Object. We are to demonſtrate, if the Line D I hath the Inclination given to a
Picture, and the Looking-Glaſs G H hath the In-
clination we have preſcribed, that the Angle
B a F, will be equal to the Angle B C D. Now to
prove this, draw the Line F I parallel to the Ho-
rizon, then the two Angles I D F and D F I, of
the Triangle I D F, are together equal to the
Angle D I E; but the Angle D F I, which is the
Inclination of the Looking-Glaſs, is equal to half the Angle D I E, leſs half the Angle I F a; and conſequently it is leſs than the Angle F D I,
by the Quantity of the whole Angle I F a: Therefore if the Angle I F a be added to the An-
gle D F I, we ſhall have the Angle D F a, equal
to the Angle F D I: Therefore the Angle
Fa B will be likewiſe equal to the Angle BCD. Which was to he demonſtrated.

In reaſoning nearly after the ſame Manner,
we demonſtrated what is mentioned concer- ning the Inclination of the Mirrour, when the
Box is inclin’d a little backwards.

213. FINIS.

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