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.

Guest Dec 18, 2019

#1**+1 **

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

\(f=\alpha+\beta+b+c=90\)

.Gadfly Dec 18, 2019