I  If two circles, say, C1 and C2 intersect with common tangent P1Q1 and intersection points A and B, then line AB bisects P1Q1 at, say, S.

II If the base of a triangle , say, Q1B inscribed in a circle is parallel to a tangent, say, P1P2 at its apex P1 then P1Q1=P1B1, and the ray from the apex to the base passing through the origin of the circle bisects the triangle into two RHT. (Constructing suitable copies of these bisecting rays during construction mars an otherwise very pretty construction, but may make the result more obvious.)

III Putting together I and II, in the given construction, lines P1Q1 and BS are parallel.

IV It follows at once that Q1B=P1S=SP2.

Construction.
A triangle congruent to Q1BP1 and rotated through 180º is constructed betwen the parallel lines at P1BS. Call a triangle with the same orientation as Q1BP1, H1, and with the same orientation as P1BS, H2. We can put a copy of H2, H2' at SP2 with apex at, say B', and the remaining space BB'S is a copy of H1, H1'. We keep alternating these copies until we reach the triangle with a vertex on R2.

P1P2=2Q1B;BR2=3Q1B; BR2/BQ1=3

Edited on October 24, 2011, 1:46 am

Posted by broll on 2011-10-24 01:10:47