III.17 Proposition III.17
From a given point to draw a straight line touching a given circle.
Proof
Let A be the external point and BCD the circle with centre E.
Join AE, and at E erect a perpendicular to AE (I.11); with
E as centre and EA as radius describe a circle AFG, meeting
the perpendicular at F. Join FA, meeting the original circle
BCD at B. Then AB is the desired tangent.
Proof: △ABE and △AFE are congruent by SAS
(EA common, EB=EF both equal to the radius of AFG,
∠AEB=∠AEF by construction); hence ∠ABE=∠AFE, which is right. So AB⊥EB, the radius at the
point of contact, and by III.18 (next) AB is tangent.
lines 74–74 in main.tex
