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Piont C lies on a circle one of whose diameter  is AB.  The bisector of angle CAB intersects BC at D and intersects the circle at E.  If BD = 25 and CD = 7, find BE.

 Dec 16, 2019
 #1
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Here's a pic of the situation :

 

 

 

Since AD  is an angle bisector.....then  DB/ DC  = AB/AC

So

25/7  = AB/ AC

AB = (25/7)AC

 

And  since AB is a diamemter,  triangle ACB  is right....and BC  = BD + DC  =  25 + 7  = 32

So

BC^2  + AC^2  = AB^2

32^2  + AC^2  = [(25/7)AC]^2

1024  = [625/49  - 1 ] AC^2

1024  = [576/49]AC^2

1024 * [49 / 576]  = AC^2

AC = sqrt (1024 * 49/ 576 ]  =   32*7 / 24  = 28/3

 

 

And triangle ACD  is also right...with hypotenuse AD  and legs CD and AC....so....

AD^2 = CD^2  + AC^2

AD^2  = 7^2  + (28/3)^2

AD^2 = 49 + 784/ 9

AD^2 = [ 441 + 784 ] / 9

AD=  sqrt [ 1225/ 9]  = 35/3

 

And by the intersecting chord theorem

 

BD * DC = AD*DE

25 * 7  = (35/3)(DE)

175 = (35/3) (DE)

DE = 175 (3) / 35  = 15

 

And  triangle DEB is also right   with   hypotenuse DB = 25   and leg DE = 15

So

DB^2  = DE^2 + BE^2

25^2  = 15^2  + BE^2

25^2 - 15^2 = BE^2

625 - 225 = BE^2

400 = BE^2

20  = BE

 

 

cool cool cool

 Dec 16, 2019

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