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Recall that tangent segments to a circle from an external point are equal in length,
=> AD=AE=5, BD=BF=3⇒AD=AE=5,BD=BF=3,
let CE=x, => CF=CE=xCE=x,⇒CF=CE=x
By the law of cosines,
BC^2=AB^2+AC^2-2*AB*AC*cosangleABC2=AB2+AC2−2⋅AB⋅AC⋅cos∠A
=> (3+x)^2=8^2+(5+x)^2-2*8*(5+x)*cos60⇒(3+x)2=82+(5+x)2−2⋅8⋅(5+x)⋅cos60
=> 9+6x+x^2=64+25+10x+x^2-2*8*(5+x)*1/2⇒9+6x+x2=64+25+10x+x2−2⋅8⋅(5+x)⋅12
=> 6x-10x+8x=64+25-40-9⇒6x−10x+8x=64+25−40−9
=> 4x=40, => x=10⇒4x=40,⇒x=10 cm
Hence, BC=BF+FC=3+x=3+10=13BC=BF+FC=3+x=3+10=13 cm
Solution 2 :
![enter image source here]()
Let O and rOandr be the center and the radius of the circle, respectively, as shown in the figure.
As AD=AE, OD=OE=r, and OAAD=AE,OD=OE=r,andOA is the common hypotenuse,
=> DeltaAOD and DeltaAOE are congruent,
=> AO bisects angleDAE, => angleOAD=60/2=30^@
=> r=AD*tan30=5*tan30=5/sqrt3
Area of ABC=|ABC|=1/2*AB*AE*sin60
=1/2*8*(5+x)*sqrt3/2=10sqrt3+2sqrt3x-----Eq(1)
|ABC|=1/2*r*(AB+AC+BC)
=1/2*5/sqrt3*(8+5+x+3+x)
=5/(2sqrt3)*(16+2x)-----Eq(2)
Eq(1)=Eq(2),
=> 10sqrt3+2sqrt3x=5/(2sqrt3)(16+2x)
=> 60+12x=80+10x,
=> 2x=20, => x=10 cm
=> BC=3+x=3+10=13 cm
