Let $\mathfrak K_1$ and $\mathfrak K_2$ be circles centered at $O_1$ and $O_2,$ respectively, meeting at the points $A$ and $B.$ Let $p$ be the line through the point $A$ meeting the circles $\mathfrak K_1$ and $\mathfrak K_2$ again at $C_1$ and $C_2.$ Assume that $A$ lies between $C_1$ and $C_2.$ Denote the intersection of the lines $C_1O_1$ and $C_2O_2$ by $D.$ Prove that the points $C_1, C_2, B$ and $D$ lie on the same circle.
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Tags: geometry proposed, geometry
jayme
17.11.2010 16:17
Dear Mathlinkers, Consider the tangents to K1, K2 resp. at C1, C2... Who want to continue? Sincerely Jean-Louis
oneplusone
18.11.2010 07:59
$\angle C_1BC_2=\angle C_1BA+\angle C_2BA=90-\angle AC_1O_1+90-\angle AC_2O_2=\angle C_1DC_2$ so $C_1C_2DB$ is cyclic.
jayme
14.12.2012 15:46
Dear Mathlinkers, 1. P, Q the second points of intersection of AO1, AO2 resp. with K1, K2 2. Consider the Morley's circle 3. After an angle chasing we are done Sincerely Jean-Louis
sayantanchakraborty
27.01.2014 11:45
Very easy Let $\angle {C_1O_1A}=2\theta, \angle {C_2O_2A}=2\alpha$ Then $\angle {C_1BA}=\theta,\angle {C_2BA}=\alpha$ ,so $\angle {C_1BC_2}=\theta+\alpha$ Also $\angle {O_1C_1A}= 90^{\circle}-\theta$ $\angle {O_2C_2A}= 90^{\circle}-\alpha$ so $\angle {C_1DC_2}= \theta+\alpha$ Hence $\angle {C_1BC_2}=\angle {C_1DC_2}=\theta+\alpha$ and so points $C_1,B,D,C_2$ lie on a circle. Maths is the doctor of science.... Sayantan..