+1459 In the diagram below, we have $\angle ABC = \angle CAB = \angle DEB=\angle BDE.$ Find $AB$ and $CD$.
+57 Well, DEB is isoceles, so EB=3. then, i don't know what to do, but i have a feeling that be and ad are angle bisectors, but i can't prove it???
+129883 Since angle BDE = angle DEB then EB = DB = 3
Using the Law of Cosines
EB^2 = DB^2 + ED^2 - 2(DB)*ED) cos BDE
3^2 = 3^2 + 2^2 - 2(3)(2) cos BDE
[ 3^2 - 3^2 -2^2] / [ -2 * 3 * 2 ] = cos BDE
1/3 = cos BDE
And angle ABD = angle BDE =angle BDA
So AD = AB
Let AE = x
AB = 2 + x
By the Law of Cosines again
AD^2 = AB^2 + DB^2 - 2(AB)(DB)cos DBA
(2 +x)^2 = (2 + x)^2 + 3^2 - 2(2+x)(3)(1/3)
-3^2 = -2(2 + x)
-9 = -4 - 2x
-5 = -2x
x = 5/2 = AE
So AB = 2 + x = 2 + 5/2 = 9/2 = 4.5
Let CD = y
So BC = 3 + y = AC (remember that angle ABC = angle CAB)
Law of Cosines again
BC^2 = AB^2 + AC^2 - 2 ( AB * AC) cos CAB (cos CAB = cos BDE)
(3 + y)^2 = 4.5^2 + (3 + y)^2 - 2 [ (3+y) (4.5) ] (1/3)
4.5^2 = 3 (3 + y)
20.25 = 9 + 3y
11.25 / 3 = y = 3.75 = CD
