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.
+121 
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
\(6^2=BF^2+CF^2-2(BF)(CF)cos(f)=BF^2+CF^2-2(BF)(CF)cos (90) \)
\(=BF^2+CF^2\)
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
\(\alpha+2b+\beta=80\)
and from the interior angles of EBC we get
\(\beta+2c+\alpha=100\)
Adding the second and third equation we get
\(2\alpha+2\beta+2b+2c=180\), giving us
\(\alpha+\beta+b+c=90\).
This fact added to the first equation results in
\(f=\alpha+\beta+b+c=90\)
