Circle Geometry - CEMC - University of Waterloo

Euclid eWorkshop #6
CIRCLE GEOMETRY
Extension
In the diagram, P AB and P CD are two secants of the same circle and they intersect at the point P outside the circle.
Prove that (P A)(P B) = (P C)(P D).
D
B
C
A
T
P
Solution
Consider △P AD and △P CB. Since both share ∠AP C we need only have one more pair of angles equal to establish
similarity. But ∠ADC and ∠ABC both are subtended by the arc AC and so are equal. Thus ∆P AD is similar to
∆P CB and using the same logic as above we arrive at (P A)(P B) = (P C)(P D).
Now imagine a series of secants passing through P , each intersecting the circle at two points but with the chord that
is within the circle getting shorter and shorter as the secant approaches the point where it intersects the circle at a
single point. In this case the secant becomes a tangent to the circle at the limiting point of intersection which we
label T . We notice that as this process takes place, P A approaches P T and P B approaches P T . Thus we have
(P A)(P B) = (P C)(P D) = (P T ) 2 . Try to prove this last statement directly using similarity and result (IV) below!
Other Important Properties of Tangents
If P is a point outside of a circle and we draw the two tangents to the circle P T and P S then the following results
follow:
I. A tangent at a point on a circle is perpendicular to the radius drawn to the point. (OT is perpendicular to P T )
II. P S = P T : Tangents to a circle from an external point are equal.
III. OP bisects the angle between the tangents. (∠T P S).
O
S
T
P
IV. Tangent Chord Theorem: Given that T A is any chord of a circle and P T is tangent to the circle at T . If C is
a point on the circle chosen to be on the side of the chord opposite to the tangent then ∠T CA = ∠P T A.
C
O
A
T
P
Proof
Since we know OT is perpendicular to T P we draw in the radii OT and OA. Since the tangent is perpendicular
to the radius, ∠P T A = 90 ◦ − ∠AT O = 1 2 (180◦ − ∠AT O − ∠OAT ) = 1 (∠AOT ) = ∠ACT .
2
CANADIAN MATHEMATICS COMPETITION 4