# A Course of Pure Geometry by E. H. Askwith By E. H. Askwith

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In the drawing we T have, RI = 9, MR = 11, and MI = 15. Let’s give RA the value of x and see what we can do. If RA is x, then MA = 11–x I since MR = 11. All radii in a circle are N M congruent so MN = 11–x also. Since MI = 15, we know that NI = 15 – (11–x) = 4 + x. Again, all radii in a circle are congruent, so TI = 4 + x also. Knowing that RI = 9 lastly gives us that RT = 9 – (x + 4) = 5–x. 5. 5 (traveling counter-clockwise). Example: In this diagram, find the radius of the circle given that the chord has length 16.

Secant-Secant Power Theorem: The product of the lengths of one secant and its external part is equal to the product of the lengths of the other secant and its external part. These four theorems concerning secants and tangents are reiterated graphically below. L # D O # Two tangents from a common exterior point are congruent – in this case, DL ≅ DY Any radius drawn to a point of tangency is perpendicular to the tangent – in this case, OL ⊥ DL and OY ⊥ DY Y H If a tangent and a secant are drawn from a common exterior point, the product of the secant’s length and the length of its external part equals the square of the length of the tangent: in this case, UE × SE = (HE) 2 [Secant-Tangent Power Theorem] # If two secants are drawn from a common point, the product of the first secant’s length and the length of its external part equals the product of the second secant’s length and the length of its external part: in this case, SC × SU = SN × SH [Secant-Secant Power Theorem] E S O # U C U S O N H 31 GEOMETRY RESOURCE DEMIDEC RESOURCES © 2001 Example: Find the measure of arc PI given that m∠P = 20°.

A triangle joining the centers of the three circles has sides 9, 11, and 15. What are the radii of the three circles? Solution: Without a picture given, we must draw one. We stop to draw three circles that R are externally tangent to each other. The two largest circles contain the triangle A side of length 15. In the drawing we T have, RI = 9, MR = 11, and MI = 15. Let’s give RA the value of x and see what we can do. If RA is x, then MA = 11–x I since MR = 11. All radii in a circle are N M congruent so MN = 11–x also.