Geometry Problem. Post your solution in the comments box below.

Level: Mathematics Education, High School, Honors Geometry, College.

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## 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)