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.
+128407 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
