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In triangle $ABC$, let the perpendicular bisector of $BC$ intersect $BC$ and $AC$ at $D$ and $E$, respectively. If $BC = 20$ and $\angle C = 15^\circ$, then find the length of $BE$.

 May 28, 2024
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                     A    E 75°

 

       B                  D                   C15°

              10                   10

 

Since ED is a  perpendicular bisector... BD / CD = BE / CE 

 

BD = CD   so   BE = CE   

 

CE  = DC / sin DEC

 

CE  = 10 / sin (75°)

 

CE = 10 / sin (30 + 45)

 

CE  =10 / [ sin30 cos 45 + sin45 cos 30 ] 

 

CE = 10 [ (1/2)(sqrt 2/2  + sqrt2/2sqrt 3/2]

 

CE =10 [ sqrt 2/4  + sqrt 6/4 ]

 

CE =  40 / [ sqrt 2 +sqrt 6]

 

CE = 40 [ sqrt 6 -sqrt 2 ] / [ 6 -2]

 

CE =  10 [ sqrt 6 -sqrt 2]  ≈ 10.35 =  BE

 

cool cool cool

 May 28, 2024
edited by CPhill  May 28, 2024

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