tag:blogger.com,1999:blog-6933544261975483399.post8640977734458814270..comments2023-03-29T05:17:23.129-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 941: Circles Tangent Externally, Tangent, Secant, Proportion, AngleAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger1125tag:blogger.com,1999:blog-6933544261975483399.post-23773008364045383652013-12-26T20:35:35.909-08:002013-12-26T20:35:35.909-08:001. Let CA and DA cut circle Q at L and M ∆COA and...1. Let CA and DA cut circle Q at L and M<br />∆COA and ∆LQA are similar ( case AA)<br />Power of C to circle Q = CE^2= CA.CL<br />So CA^2/CE^2= CA/CL= OA/OQ<br />Similarly DA^2/DF^2=DA/DM= OA/OQ<br />So CA/CE=DA/DF<br />2. Draw DN//CE<br />Since ∠ (CEF)= ∠ (DFE). ….(face the same arc EF )<br />And ∠ (CEF)= ∠ (DNF) =&gt; ∠ (DNF)= ∠ (DFN) <br />So triangle DFN is isosceles and DN=DF<br />From the result of the 1st question we have CA/DA=CE/DF=CE/DN<br />Triangles BCE and BDN are similar …( Case AA) =&gt; BC/BD=CE/DN=CE/DF=AC/AD<br />So AB is an external angle bisector of angle CAD<br />And ∠ (BAC)= ∠ (DAG)<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com