Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to enlarge the problem 941.
Sunday, December 15, 2013
Geometry Problem 941: Circles Tangent Externally, Tangent, Secant, Proportion, Angle
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1. Let CA and DA cut circle Q at L and M
ReplyDelete∆COA and ∆LQA are similar ( case AA)
Power of C to circle Q = CE^2= CA.CL
So CA^2/CE^2= CA/CL= OA/OQ
Similarly DA^2/DF^2=DA/DM= OA/OQ
So CA/CE=DA/DF
2. Draw DN//CE
Since ∠ (CEF)= ∠ (DFE). ….(face the same arc EF )
And ∠ (CEF)= ∠ (DNF) => ∠ (DNF)= ∠ (DFN)
So triangle DFN is isosceles and DN=DF
From the result of the 1st question we have CA/DA=CE/DF=CE/DN
Triangles BCE and BDN are similar …( Case AA) => BC/BD=CE/DN=CE/DF=AC/AD
So AB is an external angle bisector of angle CAD
And ∠ (BAC)= ∠ (DAG)