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In triangle $ABC,$ $\angle C = 90^\circ.$ A semicircle is constructed along side $\overline{AC}$ that is tangent to $\overline{BC}$ and $\overline{AB}.$ If the radius of the semicircle is equal to $\frac{BC}{3},$ then find $\frac{AC}{BC}.$

 Jan 16, 2024
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Angle AED = 90  = Angle BED

DC = DE

DB  = DB

Triangle DBC congrunent to triangle DBE  by HL

tan (DBC)  = DC / BC  = (BC/3) / BC   = 1/3  = tan (DBE)

 

tan (EBC)  =  AC  / BC  = 

tan (2 * DBC)   = 

 [ tan (DBE) + tan (DBC) ]  / [ 1 - tan(DBE)*tan (DBC) ] =

(1/3 + 1/3 )  /  ( 1 - (1/3)(1/3) )  =

(2/3) / ( 8/9)  =

18/24 =

3/4 =  AC / BC

 

cool cool cool

 Mar 1, 2024
edited by CPhill  Mar 10, 2024

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