In triangle ABC, BC = 6.  Point E is on AC so that angle CEB is 80 degrees.  Point D is on AB so that angle BDC is 100 degrees.  The bisectors of angles ECD and DBE intersect at F.  Find the numerical value of CF^2 + FB^2.

 Dec 18, 2019

If we show that the measure of angle f in the triangle BCF is 90 degrees, then the law of cosines

would tell us that

\(6^2=BF^2+CF^2-2(BF)(CF)cos(f)=BF^2+CF^2-2(BF)(CF)cos (90) \)



since cos(90) = 0.


But f is indeed 90 degrees:


The three interior angles of BFC add up to 180, so


\(f+\alpha +c+\beta+b=180\)


The three interior angles of DBC add up to 180, so




and from the interior angles of EBC we get




Adding the second and third equation we get


\(2\alpha+2\beta+2b+2c=180\), giving us



This fact added to the first equation results in



 Dec 18, 2019

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