Online Geometry theorems, problems, solutions, and related topics.
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.
1. Let CA and DA cut circle Q at L and M∆COA and ∆LQA are similar ( case AA)Power of C to circle Q = CE^2= CA.CLSo CA^2/CE^2= CA/CL= OA/OQSimilarly DA^2/DF^2=DA/DM= OA/OQSo CA/CE=DA/DF2. Draw DN//CESince ∠ (CEF)= ∠ (DFE). ….(face the same arc EF )And ∠ (CEF)= ∠ (DNF) => ∠ (DNF)= ∠ (DFN) So triangle DFN is isosceles and DN=DFFrom the result of the 1st question we have CA/DA=CE/DF=CE/DNTriangles BCE and BDN are similar …( Case AA) => BC/BD=CE/DN=CE/DF=AC/ADSo AB is an external angle bisector of angle CADAnd ∠ (BAC)= ∠ (DAG)