PRINCIPIA · THEOREM

Two tangents from an external point have equal length

Depends on: Tangent is perpendicular to the radius at the point of tangency, HL (hypotenuse-leg) congruence.

Statement

Let O\odot O be the circle of centre OO and radius RR, and let PP be a point outside O\odot O. From PP draw two tangents to O\odot O, touching at AA and BB. Then the two tangent lengths are equal:

PA  =  PB.|PA| \;=\; |PB|.

Furthermore, the segment OPOP joining the centre to PP bisects both symmetric angles:

APO  =  BPO,AOP  =  BOP.\angle APO \;=\; \angle BPO,\qquad \angle AOP \;=\; \angle BOP.

That is, OPOP is simultaneously the Angle bisector ⇔ equidistant from sides of APB\angle APB and of AOB\angle AOB.

Equal-tangents theorem: from an external point P of \odot O, two tangents touch at A and B ⇒ |PA| = |PB|, and OP bisects \angle APB and \angle AOB

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