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.
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
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